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Using derivatives to evaluate limits

Because differential calculus is based on the definition of the derivative, and the definition of the derivative involves a limit, there is a sense in which all of calculus rests on limits.

Introduction

Because differential calculus is based on the definition of the derivative, and the definition of the derivative involves a limit, there is a sense in which all of calculus rests on limits. In addition, the limit involved in the definition of the derivative always generates the indeterminate form \(\frac{0}{0}\). If \(f\) is a differentiable function, then in the definition \[\begin{aligned}\end{aligned}\], not only does \(h \to 0\) in the denominator, but also \((f(x+h)-f(x)) \to 0\) in the numerator, since \(f\) is continuous. Remember, saying that a limit has an indeterminate form only means that we don't yet know its value and have more work to do: indeed, limits of the form \(\frac{0}{0}\) can take on any value, as is evidenced by evaluating \(f'(x)\) for varying values of \(x\) for a function such as \(f(x) = x^2\).

We have learned many techniques for evaluating the limits that result from the derivative definition, including a large number of shortcut rules. In this section, we turn the situation upside-down: instead of using limits to evaluate derivatives, we explore how to use derivatives to evaluate certain limits.

Exploration

Using derivatives to evaluate indeterminate limits of the form \frac{0}{0}.

The idea demonstrated in Preview Activity that we can evaluate an indeterminate limit of the form \(\frac{0}{0}\) by replacing each of the numerator and denominator with their local linearizations at the point of interest can be generalized in a way that enables us to evaluate a wide range of limits. We have a function \(h(x)\) that can be written as a quotient \(h(x) = \frac{f(x)}{g(x)}\), where \(f\) and \(g\) are both differentiable at \(x=a\) and for which \(f(a) = g(a) = 0\). We would like to evaluate the indeterminate limit given by \(\lim_{x \to a} h(x)\). Figure illustrates the situation. We see that both \(f\) and \(g\) have an \(x\)-intercept at \(x = a\). Their respective tangent line approximations \(L_f\) and \(L_g\) at \(x = a\) are also shown in the figure. We can take advantage of the fact that a function and its tangent line approximation become indistinguishable as \(x \to a\).

First, let's recall that \(L_f(x) = f'(a)(x-a) + f(a)\) and \(L_g(x) = g'(a)(x-a) +g(a)\). Because \(x\) is getting arbitrarily close to \(a\) when we take the limit, we can replace \(f\) with \(L_f\) and replace \(g\) with \(L_g\), and thus we observe that \[\begin{aligned}\lim_{x \to a} \frac{f(x)}{g(x)} \amp= \lim_{x \to a} \frac{L_f(x)}{L_g(x)} \\ \amp= \lim_{x \to a} \frac{f'(a)(x-a) + f(a)}{g'(a)(x-a) + g(a)}\end{aligned}\].

Next, we remember that both \(f(a) = 0\) and \(g(a) = 0\), which is precisely what makes the original limit indeterminate. Substituting these values for \(f(a)\) and \(g(a)\) in the limit above, we now have \[\begin{aligned}\lim_{x \to a} \frac{f(x)}{g(x)} \amp= \lim_{x \to a} \frac{f'(a)(x-a)}{g'(a)(x-a)} \\ \amp= \lim_{x \to a} \frac{f'(a)}{g'(a)}\end{aligned}\], where the latter equality holds because \(\frac{x-a}{x-a} = 1\) when \(x\) is approaching (but not equal to) \(a\). Finally, we note that \(\frac{f'(a)}{g'(a)}\) is constant with respect to \(x\), and thus \[\begin{aligned}\end{aligned}\].

This result holds as long as \(g'(a)\) is not equal to zero. The formal name of the result is L'Hôpital's Rule.

Let \(f\) and \(g\) be differentiable on an open interval that includes \(x=a\), and suppose that \(f(a) = g(a) = 0\) and that \(g'(a) \neq 0\). Then \[\begin{aligned}\end{aligned}\]

For example, if we consider the limit from Preview Activity, \[\begin{aligned}\end{aligned}\], by L'Hôpital's Rule we have that \[\begin{aligned}\end{aligned}\].

Condensed — the full section is in Boelkins, Active Calculus.

Limits involving \infty

The concept of infinity, denoted \(\infty\), arises naturally in calculus, as it does in much of mathematics. It is important to note from the outset that \(\infty\) is a concept, but not a number itself. Indeed, the notion of \(\infty\) naturally invokes the idea of limits. Consider, for example, the function \(f(x) = \frac{1}{x}\), whose graph is pictured in Figure.

We note that \(x = 0\) is not in the domain of \(f\), so we may naturally wonder what happens as \(x \to 0\). As \(x \to 0^+\), we observe that \(f(x)\) increases without bound. That is, we can make the value of \(f(x)\) as large as we like by taking \(x\) closer and closer (but not equal) to 0, while keeping \(x \gt 0\). This is a good way to think about what infinity represents: a quantity is tending to infinity if there is no single number that the quantity is always less than.

Recall that the statement \(\lim_{x \to a} f(x) = L\), means that we can make \(f(x)\) as close to \(L\) as we'd like by taking \(x\) sufficiently close (but not equal) to \(a\). We now expand this notation and language to include the possibility that either \(L\) or \(a\) can be \(\infty\). For instance, for \(f(x) = \frac{1}{x}\), we now write \[\begin{aligned}\end{aligned}\], by which we mean that we can make \(\frac{1}{x}\) as large as we like by taking \(x\) sufficiently close (but not equal) to 0. In a similar way, we write \[\begin{aligned}\end{aligned}\], since we can make \(\frac{1}{x}\) as close to 0 as we'd like by taking \(x\) sufficiently large (i.e., by letting \(x\) increase without bound).

In general, the notation \(\lim_{x \to a} f(x) = \infty\) means that we can make \(f(x)\) as large as we like by taking \(x\) sufficiently close (but not equal) to \(a\), and the notation \(\lim_{x \to \infty} f(x) = L\) means that we can make \(f(x)\) as close to \(L\) as we like by taking \(x\) sufficiently large. This notation also applies to left- and right-hand limits, and to limits involving \(-\infty\). For example, returning to Figure and \(f(x) = \frac{1}{x}\), we can say that \[\begin{aligned}\end{aligned}\].

Finally, we write \[\begin{aligned}\end{aligned}\] if we can make the value of \(f(x)\) as large as we'd like by taking \(x\) sufficiently large. For example, \[\begin{aligned}\end{aligned}\].

Condensed — the full section is in Boelkins, Active Calculus.

Summary

Derivatives can be used to help us evaluate indeterminate limits of the form \(\frac{0}{0}\) through L'Hôpital's Rule, by replacing the functions in the numerator and denominator with their tangent line approximations. In particular, if \(f(a) = g(a) = 0\) and \(f\) and \(g\) are differentiable on an open interval containing \(a\), L'Hôpital's Rule tells us that \[\begin{aligned}\end{aligned}\].
When we write \(x \to \infty\), this means that \(x\) is increasing without bound. Thus, writing \(\lim_{x \to \infty} f(x) = L\) means that we can make \(f(x)\) as close to \(L\) as we like by choosing \(x\) to be sufficiently large. Similarly, \(\lim_{x \to a} f(x) = \infty\) means that we can make \(f(x)\) as large as we like by choosing \(x\) sufficiently close to \(a\).
A version of L'Hôpital's Rule also helps us evaluate indeterminate limits of the form \(\frac{\infty}{\infty}\). If \(f\) and \(g\) are differentiable and both approach zero or both approach \(\pm \infty\) as \(x \to a\) (where \(a\) is allowed to be \(\infty\)), then \[\begin{aligned}\end{aligned}\].

Practice (11)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Explain why \[\begin{aligned}\end{aligned}\] is indeterminate, and then (if possible) find the numerical value of the limit. Write to explain your thinking (e.g. did you reason algebraically, numerically, or graphically?).
Explain why \[\begin{aligned}\end{aligned}\] is indeterminate, and then (if possible) find the numerical value of the limit. Write to explain your thinking.
Explain why \[\begin{aligned}\end{aligned}\] is indeterminate, and then (if possible) find the numerical value of the limit. Write to explain your thinking.
Let \(f\) and \(g\) be differentiable functions about which the following information is known: \(f(3) = g(3) = 0\), \(f'(3) = g'(3) = 0\), \(f''(3) = -2\), and \(g''(3) = 1\). Let a new function \(h\) be given by the rule \(h(x) = \frac{f(x)}{g(x)}\). On the same set of axes, sketch possible graphs of \(f\) and \(g\) near \(x = 3\), and use the provided information to determine the value of \[\begin{aligned}\end{aligned}\].
Provide explanation to support your conclusion.

జవాబు వెల్లడి చేయండి

See the figure below for possible graphs of \(f\) and \(g\). To evaluate the limit, we can apply L'Hôpital's Rule twice. Since \(f(3) = g(3) = 0\), we know by the rule that \[\begin{aligned}\end{aligned}\]. Since \(f'(3) = g'(3) = 0\), the limit is still indeterminate, and we can apply the rule again to get \[\begin{aligned}\end{aligned}\]. Hence, \(\lim_{x \to 3} h(x) = -2\).
Find all vertical and horizontal asymptotes of the function \[\begin{aligned}\end{aligned}\], where \(a\), \(b\), and \(c\) are distinct, arbitrary constants. In addition, state all values of \(x\) for which \(R\) is not continuous. Sketch a possible graph of \(R\), clearly labeling the values of \(a\), \(b\), and \(c\).

జవాబు వెల్లడి చేయండి

First, we note that since both the numerator and denominator of \(R\) are quadratic, the limit as \(x \to \infty\) can be found by taking the quotient of the leading coefficients. In particular, \[\begin{aligned}\end{aligned}\] and thus the horizontal asymptote of \(R\) is \(y = \frac{3}{5}\).
Next, since \(b\) and \(c\) are distinct, we see that the numerator of \(R\) is zero (and the denominator is not) when \(x = b\), and thus \(R(b) = 0\). In addition, the denominator is zero (and the numerator is not) when \(x = c\), and this results in there being a vertical asymptote at \(x = c\).
Finally, we observe that \(x = a\) makes both the numerator and denominator zero. Since \(R(a)\) is thus not defined, the graph of \(R\) has either a hole or a vertical asymptote at \(x = a\). To determine which, we evaluate the limit and see that \[\begin{aligned}\end{aligned}\]. This tells us that \(R\) has a hole at the point \((a, \frac{3(a-b)}{5(a-c)})\). Moreover, \(R\) is not continuous at \(x = a\) and \(x = c\).
A possible graph of \(R\) is shown in the following figure.
Consider the function \(g(x) = x^{2x}\), which is defined for all \(x \gt 0\). Observe that \(\lim_{x \to 0^+} g(x)\) is indeterminate due to its form of \(0^0\). (Think about how we know that \(0^k = 0\) for all \(k \gt 0\), while \(b^0 = 1\) for all \(b \ne 0\), but that neither rule can apply to \(0^0\).)
1. Let \(h(x) = \ln(g(x))\). Explain why \(h(x) = 2x \ln(x)\).
2. Next, explain why it is equivalent to write \(h(x) = \frac{2\ln(x)}{\frac{1}{x}}\).
3. Use L'Hôpital's Rule and your work in (b) to compute \(\lim_{x \to 0^+} h(x)\).
4. Based on the value of \(\lim_{x \to 0^+} h(x)\), determine \(\lim_{x \to 0^+} g(x)\).
జవాబు వెల్లడి చేయండి
1. Since \(h(x) = \ln(g(x))\) and we are given that \(g(x) = x^{2x}\), it follows that \[\begin{aligned}\end{aligned}\].
2. Since \(x = (x^{-1})^{-1} = \frac{1}{\frac{1}{x}}\), we can also write \[\begin{aligned}\end{aligned}\].
3. Since \(\ln(x) \to -\infty\) as \(x \to 0^+\) and \(\frac{1}{x} \to +\infty\) as \(x \to 0^+\), we see that \[\begin{aligned}\end{aligned}\] is of an indeterminate form to which we can apply L'Hôpital's Rule. Doing so, \[\begin{aligned}\end{aligned}\] Simplifying algebraically, \[\begin{aligned}\end{aligned}\] Hence, it follows that \(\lim_{x \to 0^+} h(x) = 0\).
4. Having shown that \(\lim_{x \to 0^+} h(x) = 0\) and recalling that \(h(x) = \ln(g(x))\), we know that as \(x \to 0^+\), the natural log of \(g(x)\) tends to \(0\). Since \(\ln(1) = 0\), it must be the case that \(g(x)\) is tending to \(1\), and thus \[\begin{aligned}\end{aligned}\]
Recall we say that function \(g\) dominates function \(f\) provided that \(\lim_{x \to \infty} f(x) = \infty\), \(\lim_{x \to \infty} g(x) = \infty\), and \(\lim_{x \to \infty} \frac{f(x)}{g(x)} = 0\).
1. Which function dominates the other: \(\ln(x)\) or \(\sqrt{x}\)?
2. Which function dominates the other: \(\ln(x)\) or \(\sqrt[n]{x}\)? (\(n\) can be any positive integer)
3. Explain why \(e^x\) will dominate any polynomial function.
4. Explain why \(x^n\) will dominate \(\ln(x)\) for any positive integer \(n\).
5. Give any example of two nonlinear functions such that neither dominates the other.
జవాబు వెల్లడి చేయండి
1. We note that both \(\ln(x)\) or \(\sqrt{x}\) increase without bound as \(x \to \infty\), so we consider \[\begin{aligned}\end{aligned}\]. By L'Hôpital's Rule and some simplifying algebra, \[\begin{aligned}\end{aligned}\]. Thus, it follows that \[\begin{aligned}\end{aligned}\], and hence \(\sqrt{x}\) dominates \(\ln(x)\).
2. Using a similar argument to our work in \((a)\), we can show that \(\sqrt[n]{x}\) dominates \(\ln(x)\) for any positive integer \(n\). In particular, by L'Hôpital's Rule and some simplifying algebra, \[\begin{aligned}\end{aligned}\]. Since \(\frac{x^{(n-1)/n}}{x} = x^{(n-1)/n - 1} = x^{(n-1-n)/n} = x^{-1/n}\), it follows that \[\begin{aligned}\end{aligned}\] we've shown that \(\sqrt[n]{x}\) dominates \(\ln(x)\) for any positive integer \(n\).
3. To see why \(e^x\) will dominate any polynomial function, consider \(p(x) = x^4\). If we consider the limit of their quotient and apply LHR once, we see that \[\begin{aligned}\end{aligned}\] By repeated application of LHR, the numerator (regardless of what polynomial we start with) will eventually be simply a constant (after \(n\) applications of LHR), and thus with \(e^x\) still in the denominator, the overall limit will be \(0\).
4. Here we naturally consider the limit of the quotient of \(\ln(x)\) and \(x^n\), where \(n \ge 1\). Since both functions increase without bound, we can apply LHR. Doing so along with some simplifying algebra, we see \[\begin{aligned}\end{aligned}\] and thus \(x^n\) indeed dominates.
5. Consider \(f(x) = 3x^2 + 1\) and \(g(x) = -0.5x^2 + 5x - 2\). Since these functions are polynomials of the same degree, we know \[\begin{aligned}\end{aligned}\].
Let \(f\) and \(g\) be differentiable functions about which the following information is known: \(f(3) = g(3) = 0\), \(f'(3) = g'(3) = 0\), \(f''(3) = -2\), and \(g''(3) = 1\). Let a new function \(h\) be given by the rule \(h(x) = \frac{f(x)}{g(x)}\). On the same set of axes, sketch possible graphs of \(f\) and \(g\) near \(x = 3\), and use the provided information to determine the value of \[\begin{aligned}\end{aligned}\].
Provide explanation to support your conclusion.

జవాబు వెల్లడి చేయండి

See the figure below for possible graphs of \(f\) and \(g\). To evaluate the limit, we can apply L'Hôpital's Rule twice. Since \(f(3) = g(3) = 0\), we know by the rule that \[\begin{aligned}\end{aligned}\]. Since \(f'(3) = g'(3) = 0\), the limit is still indeterminate, and we can apply the rule again to get \[\begin{aligned}\end{aligned}\]. Hence, \(\lim_{x \to 3} h(x) = -2\).
Find all vertical and horizontal asymptotes of the function \[\begin{aligned}\end{aligned}\], where \(a\), \(b\), and \(c\) are distinct, arbitrary constants. In addition, state all values of \(x\) for which \(R\) is not continuous. Sketch a possible graph of \(R\), clearly labeling the values of \(a\), \(b\), and \(c\).

జవాబు వెల్లడి చేయండి

First, we note that since both the numerator and denominator of \(R\) are quadratic, the limit as \(x \to \infty\) can be found by taking the quotient of the leading coefficients. In particular, \[\begin{aligned}\end{aligned}\] and thus the horizontal asymptote of \(R\) is \(y = \frac{3}{5}\).
Next, since \(b\) and \(c\) are distinct, we see that the numerator of \(R\) is zero (and the denominator is not) when \(x = b\), and thus \(R(b) = 0\). In addition, the denominator is zero (and the numerator is not) when \(x = c\), and this results in there being a vertical asymptote at \(x = c\).
Finally, we observe that \(x = a\) makes both the numerator and denominator zero. Since \(R(a)\) is thus not defined, the graph of \(R\) has either a hole or a vertical asymptote at \(x = a\). To determine which, we evaluate the limit and see that \[\begin{aligned}\end{aligned}\]. This tells us that \(R\) has a hole at the point \((a, \frac{3(a-b)}{5(a-c)})\). Moreover, \(R\) is not continuous at \(x = a\) and \(x = c\).
A possible graph of \(R\) is shown in the following figure.
Consider the function \(g(x) = x^{2x}\), which is defined for all \(x \gt 0\). Observe that \(\lim_{x \to 0^+} g(x)\) is indeterminate due to its form of \(0^0\). (Think about how we know that \(0^k = 0\) for all \(k \gt 0\), while \(b^0 = 1\) for all \(b \ne 0\), but that neither rule can apply to \(0^0\).)
1. Let \(h(x) = \ln(g(x))\). Explain why \(h(x) = 2x \ln(x)\).
2. Next, explain why it is equivalent to write \(h(x) = \frac{2\ln(x)}{\frac{1}{x}}\).
3. Use L'Hôpital's Rule and your work in (b) to compute \(\lim_{x \to 0^+} h(x)\).
4. Based on the value of \(\lim_{x \to 0^+} h(x)\), determine \(\lim_{x \to 0^+} g(x)\).
జవాబు వెల్లడి చేయండి
1. Since \(h(x) = \ln(g(x))\) and we are given that \(g(x) = x^{2x}\), it follows that \[\begin{aligned}\end{aligned}\].
2. Since \(x = (x^{-1})^{-1} = \frac{1}{\frac{1}{x}}\), we can also write \[\begin{aligned}\end{aligned}\].
3. Since \(\ln(x) \to -\infty\) as \(x \to 0^+\) and \(\frac{1}{x} \to +\infty\) as \(x \to 0^+\), we see that \[\begin{aligned}\end{aligned}\] is of an indeterminate form to which we can apply L'Hôpital's Rule. Doing so, \[\begin{aligned}\end{aligned}\] Simplifying algebraically, \[\begin{aligned}\end{aligned}\] Hence, it follows that \(\lim_{x \to 0^+} h(x) = 0\).
4. Having shown that \(\lim_{x \to 0^+} h(x) = 0\) and recalling that \(h(x) = \ln(g(x))\), we know that as \(x \to 0^+\), the natural log of \(g(x)\) tends to \(0\). Since \(\ln(1) = 0\), it must be the case that \(g(x)\) is tending to \(1\), and thus \[\begin{aligned}\end{aligned}\]
Recall we say that function \(g\) dominates function \(f\) provided that \(\lim_{x \to \infty} f(x) = \infty\), \(\lim_{x \to \infty} g(x) = \infty\), and \(\lim_{x \to \infty} \frac{f(x)}{g(x)} = 0\).
1. Which function dominates the other: \(\ln(x)\) or \(\sqrt{x}\)?
2. Which function dominates the other: \(\ln(x)\) or \(\sqrt[n]{x}\)? (\(n\) can be any positive integer)
3. Explain why \(e^x\) will dominate any polynomial function.
4. Explain why \(x^n\) will dominate \(\ln(x)\) for any positive integer \(n\).
5. Give any example of two nonlinear functions such that neither dominates the other.
జవాబు వెల్లడి చేయండి
1. We note that both \(\ln(x)\) or \(\sqrt{x}\) increase without bound as \(x \to \infty\), so we consider \[\begin{aligned}\end{aligned}\]. By L'Hôpital's Rule and some simplifying algebra, \[\begin{aligned}\end{aligned}\]. Thus, it follows that \[\begin{aligned}\end{aligned}\], and hence \(\sqrt{x}\) dominates \(\ln(x)\).
2. Using a similar argument to our work in \((a)\), we can show that \(\sqrt[n]{x}\) dominates \(\ln(x)\) for any positive integer \(n\). In particular, by L'Hôpital's Rule and some simplifying algebra, \[\begin{aligned}\end{aligned}\]. Since \(\frac{x^{(n-1)/n}}{x} = x^{(n-1)/n - 1} = x^{(n-1-n)/n} = x^{-1/n}\), it follows that \[\begin{aligned}\end{aligned}\] we've shown that \(\sqrt[n]{x}\) dominates \(\ln(x)\) for any positive integer \(n\).
3. To see why \(e^x\) will dominate any polynomial function, consider \(p(x) = x^4\). If we consider the limit of their quotient and apply LHR once, we see that \[\begin{aligned}\end{aligned}\] By repeated application of LHR, the numerator (regardless of what polynomial we start with) will eventually be simply a constant (after \(n\) applications of LHR), and thus with \(e^x\) still in the denominator, the overall limit will be \(0\).
4. Here we naturally consider the limit of the quotient of \(\ln(x)\) and \(x^n\), where \(n \ge 1\). Since both functions increase without bound, we can apply LHR. Doing so along with some simplifying algebra, we see \[\begin{aligned}\end{aligned}\] and thus \(x^n\) indeed dominates.
5. Consider \(f(x) = 3x^2 + 1\) and \(g(x) = -0.5x^2 + 5x - 2\). Since these functions are polynomials of the same degree, we know \[\begin{aligned}\end{aligned}\].

Symbols used here

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

Euler's number
2.71828…, the base whose exponential is its own derivative.

imaginary unit
i² = −1.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\neq

not equal
The two sides are different.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

y',\ y''

first and second derivative of y
Prime notation for derivatives with respect to x (or t).

C,\ C_1,\ C_2

arbitrary constants
Constants of integration fixed by initial conditions.

How to: Using derivatives to evaluate limits

How can derivatives be used to help us evaluate indeterminate limits of the form \frac{0}{0}?
What does it mean to say that \lim_{x \to \infty} f(x) = L and \lim_{x \to a} f(x) = \infty?
How can derivatives assist us in evaluating indeterminate limits of the form \frac{\infty}{\infty}?

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

మీ సొంత ప్రయత్నించండి

Parts of this page are adapted from Boelkins, Active Calculus (CC BY-SA 4.0). Condensed and re-explained here; errors are ours.

ఇంకా Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests