maths.free › Calculus › 12. Introduction to Calculus › Finding Limits: Properties of Limits

Finding Limits: Properties of Limits

Find the limit of a sum, a difference, and a product.

Finding the Limit of a Sum, a Difference, and a Product

Graphing a function or exploring a table of values to determine a limit can be cumbersome and time-consuming. When possible, it is more efficient to use the properties of limits, which is a collection of theorems for finding limits.

Knowing the properties of limits allows us to compute limits directly. We can add, subtract, multiply, and divide the limits of functions as if we were performing the operations on the functions themselves to find the limit of the result. Similarly, we can find the limit of a function raised to a power by raising the limit to that power. We can also find the limit of the root of a function by taking the root of the limit. Using these operations on limits, we can find the limits of more complex functions by finding the limits of their simpler component functions.

Properties of Limits

Let \(a,\ k,\ A,\) and \(B\) represent real numbers, and \(f\) and \(g\) be functions, such that \(\underset{x\to a}{\lim }f(x)=A\) and \(\underset{x\to a}{\lim }g(x)=B.\) For limits that exist and are finite, the properties of limits are summarized in

Constant, k	\(\underset{x\to a}{\lim }k=k\)
Constant times a function	\(\underset{x\to a}{\lim }[k⋅f(x)]=k\underset{x\to a}{\lim }f(x)=kA\)
Sum of functions	\(\underset{x\to a}{\lim }[f(x)+g(x)]=\underset{x\to a}{\lim }f(x)+\underset{x\to a}{\lim }g(x)=A+B\)
Difference of functions	\(\underset{x\to a}{\lim }[f(x)-g(x)]=\underset{x\to a}{\lim }f(x)-\underset{x\to a}{\lim }g(x)=A-B\)
Product of functions	\(\underset{x\to a}{\lim }[f(x)⋅g(x)]=\underset{x\to a}{\lim }f(x)⋅\underset{x\to a}{\lim }g(x)=A⋅B\)
Quotient of functions	\(\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\frac{\underset{x\to a}{\lim }f(x)}{\underset{x\to a}{\lim }g(x)}=\frac{A}{B},B\ne 0\)
Function raised to an exponent	\(\underset{x\to a}{\lim }{[f(x)]}^{n}={[\underset{x\to a}{\lim }f(x)]}^{n}={A}^{n},\) where \(n\) is a positive integer
nth root of a function, where n is a positive integer	\(\underset{x\to a}{\lim }\sqrt[n]{f(x)}=\sqrt[n]{\underset{x\to a}{\lim }[f(x)]}=\sqrt[n]{A}\)
Polynomial function	\(\underset{x\to a}{\lim }p(x)=p(a)\)

Example

Try it.

Evaluate \(\underset{x\to 3}{\lim }(2x+5).\)

Solution

\[\begin{array}{ll}\underset{x\to 3}{\lim }(2x+5)=\underset{x\to 3}{\lim }(2x)+\underset{x\to 3}{\lim }(5) & \text{Sum of functions property} \\ =\underset{x\to 3}{2\lim }(x)+\underset{x\to 3}{\lim }(5) & \text{Constant times a function property} \\ =2(3)+5 & \text{Evaluate} \\ =11 & \end{array}\]

Finding the Limit of a Polynomial

Not all functions or their limits involve simple addition, subtraction, or multiplication. Some may include polynomials. Recall that a polynomial is an expression consisting of the sum of two or more terms, each of which consists of a constant and a variable raised to a nonnegative integral power. To find the limit of a polynomial function, we can find the limits of the individual terms of the function, and then add them together. Also, the limit of a polynomial function as \(x\) approaches \(a\) is equivalent to simply evaluating the function for \(a\).

Example

Try it.

Evaluate \(\underset{x\to 3}{\lim }(5{x}^{2}).\)

Solution

\[\begin{array}{ll}\underset{x\to 3}{\lim }(5{x}^{2})=5\underset{x\to 3}{\lim }({x}^{2}) & \text{Constant times a function property} \\ =5({3}^{2}) & \text{Function raised to an exponent property} \\ =45 & \end{array}\]

Example

Try it.

Evaluate \(\underset{x\to 5}{\lim }(2{x}^{3}-3x+1).\)

Solution

\[\begin{array}{ll}\underset{x\to 5}{\lim }(2{x}^{3}-3x+1)=\underset{x\to 5}{\lim }(2{x}^{3})-\underset{x\to 5}{\lim }(3x)+\underset{x\to 5}{\lim }(1) & \text{Sum of functions} \\ =\underset{x\to 5}{2\lim }({x}^{3})-\underset{x\to 5}{3\lim }(x)+\underset{x\to 5}{\lim }(1) & \text{Constant times a function} \\ =2({5}^{3})-3(5)+1 & \text{Function raised to an exponent} \\ =236 & \text{Evaluate}\end{array}\]

Finding the Limit of a Power or a Root

When a limit includes a power or a root, we need another property to help us evaluate it. The square of the limit of a function equals the limit of the square of the function; the same goes for higher powers. Likewise, the square root of the limit of a function equals the limit of the square root of the function; the same holds true for higher roots.

Example

Try it.

Evaluate \(\underset{x\to 2}{\lim }{(3x+1)}^{5}.\)

Solution

We will take the limit of the function as \(x\) approaches 2 and raise the result to the 5^th power.

\[\begin{array}{l}\underset{x\to 2}{\lim }{(3x+1)}^{5}={(\underset{x\to 2}{\lim }(3x+1))}^{5} \\ ={(3(2)+1)}^{5} \\ ={7}^{5} \\ =\text{16,807}\end{array}\]

Finding the Limit of a Quotient

Finding the limit of a function expressed as a quotient can be more complicated. We often need to rewrite the function algebraically before applying the properties of a limit. If the denominator evaluates to 0 when we apply the properties of a limit directly, we must rewrite the quotient in a different form. One approach is to write the quotient in factored form and simplify.

Example

Try it.

Evaluate \(\underset{x\to 2}{\lim }(\frac{{x}^{2}-6x+8}{x-2}).\)

Solution

Factor where possible, and simplify.

\[\begin{array}{ll}\underset{x\to 2}{\lim }(\frac{{x}^{2}-6x+8}{x-2})=\underset{x\to 2}{\lim }(\frac{(x-2)(x-4)}{x-2}) & \text{Factor the numerator}. \\ =\underset{x\to 2}{\lim }(\frac{(x-2)(x-4)}{x-2}) & \text{Cancel the common factors}. \\ =\underset{x\to 2}{\lim }(x-4) & \text{Evaluate}. \\ =2-4=-2 & \end{array}\]

Example

Try it.

Evaluate \(\underset{x\to 5}{\lim }(\frac{\frac{1}{x}-\frac{1}{5}}{x-5}).\)

Solution

Find the LCD for the denominators of the two terms in the numerator, and convert both fractions to have the LCD as their denominator.

Condensed — the full section is in OpenStax Precalculus 2e.

Key Concepts

The properties of limits can be used to perform operations on the limits of functions rather than the functions themselves. See .
The limit of a polynomial function can be found by finding the sum of the limits of the individual terms. See and .
The limit of a function that has been raised to a power equals the same power of the limit of the function. Another method is direct substitution. See .
The limit of the root of a function equals the corresponding root of the limit of the function.
One way to find the limit of a function expressed as a quotient is to write the quotient in factored form and simplify. See .
Another method of finding the limit of a complex fraction is to find the LCD. See .
A limit containing a function containing a root may be evaluated using a conjugate. See .
The limits of some functions expressed as quotients can be found by factoring. See .
One way to evaluate the limit of a quotient containing absolute values is by using numeric evidence. Setting it up piecewise can also be useful. See .

Practice (40)

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

Evaluate \(\underset{x\to 3}{\lim }(2x+5).\)

جواب کھوليں

\[\begin{array}{ll}\underset{x\to 3}{\lim }(2x+5)=\underset{x\to 3}{\lim }(2x)+\underset{x\to 3}{\lim }(5) & \text{Sum of functions property} \\ =\underset{x\to 3}{2\lim }(x)+\underset{x\to 3}{\lim }(5) & \text{Constant times a function property} \\ =2(3)+5 & \text{Evaluate} \\ =11 & \end{array}\]
Evaluate the following limit: \(\underset{x\to -12}{\lim }(-2x+2).\)

جواب کھوليں

26
Evaluate \(\underset{x\to 3}{\lim }(5{x}^{2}).\)

جواب کھوليں

\[\begin{array}{ll}\underset{x\to 3}{\lim }(5{x}^{2})=5\underset{x\to 3}{\lim }({x}^{2}) & \text{Constant times a function property} \\ =5({3}^{2}) & \text{Function raised to an exponent property} \\ =45 & \end{array}\]
Evaluate \(\underset{x\to 4}{\lim }({x}^{3}-5).\)

جواب کھوليں

59
Evaluate \(\underset{x\to 5}{\lim }(2{x}^{3}-3x+1).\)

جواب کھوليں

\[\begin{array}{ll}\underset{x\to 5}{\lim }(2{x}^{3}-3x+1)=\underset{x\to 5}{\lim }(2{x}^{3})-\underset{x\to 5}{\lim }(3x)+\underset{x\to 5}{\lim }(1) & \text{Sum of functions} \\ =\underset{x\to 5}{2\lim }({x}^{3})-\underset{x\to 5}{3\lim }(x)+\underset{x\to 5}{\lim }(1) & \text{Constant times a function} \\ =2({5}^{3})-3(5)+1 & \text{Function raised to an exponent} \\ =236 & \text{Evaluate}\end{array}\]
Evaluate the following limit: \(\underset{x\to -1}{\lim }({x}^{4}-4{x}^{3}+5).\)

جواب کھوليں

10
Evaluate \(\underset{x\to 2}{\lim }{(3x+1)}^{5}.\)

جواب کھوليں

We will take the limit of the function as \(x\) approaches 2 and raise the result to the 5^th power.
\[\begin{array}{l}\underset{x\to 2}{\lim }{(3x+1)}^{5}={(\underset{x\to 2}{\lim }(3x+1))}^{5} \\ ={(3(2)+1)}^{5} \\ ={7}^{5} \\ =\text{16,807}\end{array}\]
Evaluate the following limit: \(\underset{x\to -4}{\lim }{(10x+36)}^{3}.\)

جواب کھوليں

\(-64\)
Evaluate \(\underset{x\to 2}{\lim }(\frac{{x}^{2}-6x+8}{x-2}).\)

جواب کھوليں

Factor where possible, and simplify.
\[\begin{array}{ll}\underset{x\to 2}{\lim }(\frac{{x}^{2}-6x+8}{x-2})=\underset{x\to 2}{\lim }(\frac{(x-2)(x-4)}{x-2}) & \text{Factor the numerator}. \\ =\underset{x\to 2}{\lim }(\frac{(x-2)(x-4)}{x-2}) & \text{Cancel the common factors}. \\ =\underset{x\to 2}{\lim }(x-4) & \text{Evaluate}. \\ =2-4=-2 & \end{array}\]
Evaluate the following limit: \(\underset{x\to 7}{\lim }(\frac{{x}^{2}-11x+28}{7-x}).\)

جواب کھوليں

\(-3\)
Evaluate \(\underset{x\to 5}{\lim }(\frac{\frac{1}{x}-\frac{1}{5}}{x-5}).\)

جواب کھوليں

Find the LCD for the denominators of the two terms in the numerator, and convert both fractions to have the LCD as their denominator.
Evaluate \(\underset{x\to -5}{\lim }(\frac{\frac{1}{5}+\frac{1}{x}}{10+2x}).\)

جواب کھوليں

\(-\frac{1}{50}\)
Evaluate \(\underset{x\to 0}{\lim }(\frac{\sqrt{25-x}-5}{x}).\)

جواب کھوليں

\(\begin{array}{ll}\underset{x\to 0}{\lim }(\frac{\sqrt{25-x}-5}{x})=\underset{x\to 0}{\lim }(\frac{(\sqrt{25-x}-5)}{x}⋅\frac{(\sqrt{25-x}+5)}{(\sqrt{25-x}+5)}) & \text{Multiply numerator and denominator by the conjugate}. \\ =\underset{x\to 0}{\lim }(\frac{(25-x)-25}{x(\sqrt{25-x}+5)}) & \text{Multiply: }(\sqrt{25-x}-5)⋅(\sqrt{25-x}+5)=(25-x)-25. \\ =\underset{x\to 0}{\lim }(\frac{-x}{x(\sqrt{25-x}+5)}) & \text{Combine like terms}. \\ =\underset{x\to 0}{\lim }(\frac{-x}{x(\sqrt{25-x}+5)}) & \text{Simplify }\frac{-x}{x}=-1. \\ =\frac{-1}{\sqrt{25-0}+5} & \text{Evaluate}. \\ =\frac{-1}{5+5}=-\frac{1}{10} & \end{array}\)
Evaluate the following limit: \(\underset{h\to 0}{\lim }(\frac{\sqrt{16-h}-4}{h}).\)

جواب کھوليں

\(-\frac{1}{8}\)
Evaluate \(\underset{x\to 4}{\lim }(\frac{4-x}{\sqrt{x}-2}).\)

جواب کھوليں

\[\begin{array}{ll}\underset{x\to 4}{\lim }(\frac{4-x}{\sqrt{x}-2})=\underset{x\to 4}{\lim }(\frac{(2+\sqrt{x})(2-\sqrt{x})}{\sqrt{x}-2}) & \text{Factor.} \\ =\underset{x\to 4}{\lim }(\frac{(2+\sqrt{x})(2-\sqrt{x})}{-(2-\sqrt{x})}) & \text{Factor }-1\text{ out of the denominator}\text{. Simplify}. \\ =\underset{x\to 4}{\lim }-(2+\sqrt{x}) & \text{Evaluate}. \\ =-(2+\sqrt{4}) & \\ =-4 & \end{array}\]
Evaluate the following limit: \(\underset{x\to 3}{\lim }(\frac{x-3}{\sqrt{x}-\sqrt{3}}).\)

جواب کھوليں

\(2\sqrt{3}\)
Evaluate \(\underset{x\to 7}{\lim }\frac{|x-7|}{x-7}.\)

جواب کھوليں

The function is undefined at \(x=7,\) so we will try values close to 7 from the left and the right.
Left-hand limit: \(\frac{|6.9-7|}{6.9-7}=\frac{|6.99-7|}{6.99-7}=\frac{|6.999-7|}{6.999-7}=-1\)
Right-hand limit: \(\frac{|7.1-7|}{7.1-7}=\frac{|7.01-7|}{7.01-7}=\frac{|7.001-7|}{7.001-7}=1\)
Since the left- and right-hand limits are not equal, there is no limit.
Evaluate \(\underset{x\to {6}^{+}}{\lim }\frac{6-x}{|x-6|}.\)

جواب کھوليں

\(-1\)
Give an example of a type of function \(f\) whose limit, as \(x\) approaches \(a,\) is \(f(a).\)

جواب کھوليں

If \(f\) is a polynomial function, the limit of a polynomial function as \(x\) approaches \(a\) will always be \(f(a).\)
When direct substitution is used to evaluate the limit of a rational function as \(x\) approaches \(a\) and the result is \(f(a)=\frac{0}{0},\) does this mean that the limit of \(f\) does not exist?
What does it mean to say the limit of \(f(x),\) as \(x\) approaches \(c,\) is undefined?

جواب کھوليں

It could mean either (1) the values of the function increase or decrease without bound as \(x\) approaches \(c,\) or (2) the left and right-hand limits are not equal.
\(\underset{x\to 0}{\lim }(3)\)
\(\underset{x\to 2}{\lim }(\frac{-5x}{{x}^{2}-1})\)

جواب کھوليں

\(\frac{-10}{3}\)
\(\underset{x\to 2}{\lim }(\frac{{x}^{2}-5x+6}{x+2})\)
\(\underset{x\to 3}{\lim }(\frac{{x}^{2}-9}{x-3})\)

جواب کھوليں

6
\(\underset{x\to -1}{\lim }(\frac{{x}^{2}-2x-3}{x+1})\)
\(\underset{x\to \frac{3}{2}}{\lim }(\frac{6{x}^{2}-17x+12}{2x-3})\)

جواب کھوليں

\(\frac{1}{2}\)
\(\underset{x\to -\frac{7}{2}}{\lim }(\frac{8{x}^{2}+18x-35}{2x+7})\)
\(\underset{x\to 3}{\lim }(\frac{{x}^{2}-9}{{x}^{2}-5x+6})\)

جواب کھوليں

6
\(\underset{x\to -3}{\lim }(\frac{-7{x}^{4}-21{x}^{3}}{-12{x}^{4}+108{x}^{2}})\)
\(\underset{x\to 3}{\lim }(\frac{{x}^{2}+2x-3}{x-3})\)

جواب کھوليں

does not exist
\(\underset{h\to 0}{\lim }(\frac{{(3+h)}^{3}-27}{h})\)
\(\underset{h\to 0}{\lim }(\frac{{(2-h)}^{3}-8}{h})\)

جواب کھوليں

\(-12\)
\(\underset{h\to 0}{\lim }(\frac{{(h+3)}^{2}-9}{h})\)
\(\underset{h\to 0}{\lim }(\frac{\sqrt{5-h}-\sqrt{5}}{h})\)

جواب کھوليں

\(-\frac{\sqrt{5}}{10}\)
\(\underset{x\to 0}{\lim }(\frac{\sqrt{3-x}-\sqrt{3}}{x})\)
\(\underset{x\to 9}{\lim }(\frac{{x}^{2}-81}{3-\sqrt{x}})\)

جواب کھوليں

\(-108\)
\(\underset{x\to 1}{\lim }(\frac{\sqrt{x}-{x}^{2}}{1-\sqrt{x}})\)
\(\underset{x\to 0}{\lim }(\frac{x}{\sqrt{1+2x}-1})\)

جواب کھوليں

1
\(\underset{x\to \frac{1}{2}}{\lim }(\frac{{x}^{2}-\frac{1}{4}}{2x-1})\)

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.

\pm

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

\neq

not equal
The two sides are different.

\pi

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

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

\infty

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

\sin,\ \cos,\ \tan

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

\log_b x,\ \ln x

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

\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: Finding Limits: Properties of Limits

Find the limit of a sum, a difference, and a product.
Find the limit of a polynomial.
Find the limit of a power or a root.
Find the limit of a quotient.
Use the properties of limits to break up the polynomial into individual terms.
Find the limits of the individual terms.
Add the limits together.
Alternatively, evaluate the function for

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.

آپ کے اپنے کو کوشش کریں

Parts of this page are adapted from OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

میں زیادہ 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