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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.

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 5th 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.
  • The limit of a polynomial function can be found by finding the sum of the limits of the individual terms.
  • 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.
  • 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.
  • Another method of finding the limit of a complex fraction is to find the LCD.
  • A limit containing a function containing a root may be evaluated using a conjugate.
  • The limits of some functions expressed as quotients can be found by factoring.
  • 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.

Haddaad Ma jiro xisaabis oo ku nool midkan, laakiin qaybaha ka mid ah waa la xisaabi karaa. Ku raaxayso mid ka hooseeya, ama ku qor adiga kuu gaar ah.

Tababar (40)

Ku day mid kasta oo ka mid ah oo ku saabsan buugaag hore. Muuji jawaabta si aad u hubiso; kuwa la xaqiijiyay waxaa la furi karaa in xalka ah tallaabo kasta.

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

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

  2. Evaluate the following limit: \(\underset{x\to -12}{\lim }(-2x+2).\)

    Jawaabta muuji

    26

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

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

  4. Evaluate \(\underset{x\to 4}{\lim }({x}^{3}-5).\)

    Jawaabta muuji

    59

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

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

  6. Evaluate the following limit: \(\underset{x\to -1}{\lim }({x}^{4}-4{x}^{3}+5).\)

    Jawaabta muuji

    10

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

    Jawaabta muuji

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

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

  8. Evaluate the following limit: \(\underset{x\to -4}{\lim }{(10x+36)}^{3}.\)

    Jawaabta muuji

    \(-64\)

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

    Jawaabta muuji

    Factor where possible, and simplify.

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

  10. Evaluate the following limit: \(\underset{x\to 7}{\lim }(\frac{{x}^{2}-11x+28}{7-x}).\)

    Jawaabta muuji

    \(-3\)

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

    Jawaabta muuji

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

  12. Evaluate \(\underset{x\to -5}{\lim }(\frac{\frac{1}{5}+\frac{1}{x}}{10+2x}).\)

    Jawaabta muuji

    \(-\frac{1}{50}\)

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

    Jawaabta muuji

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

  14. Evaluate the following limit: \(\underset{h\to 0}{\lim }(\frac{\sqrt{16-h}-4}{h}).\)

    Jawaabta muuji

    \(-\frac{1}{8}\)

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

    Jawaabta muuji

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

  16. Evaluate the following limit: \(\underset{x\to 3}{\lim }(\frac{x-3}{\sqrt{x}-\sqrt{3}}).\)

    Jawaabta muuji

    \(2\sqrt{3}\)

  17. Evaluate \(\underset{x\to 7}{\lim }\frac{|x-7|}{x-7}.\)

    Jawaabta muuji

    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.

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

    Jawaabta muuji

    \(-1\)

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

    Jawaabta muuji

    If \(f\) is a polynomial function, the limit of a polynomial function as \(x\) approaches \(a\) will always be \(f(a).\)

  20. 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?

  21. What does it mean to say the limit of \(f(x),\) as \(x\) approaches \(c,\) is undefined?

    Jawaabta muuji

    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.

  22. \(\underset{x\to 0}{\lim }(3)\)

  23. \(\underset{x\to 2}{\lim }(\frac{-5x}{{x}^{2}-1})\)

    Jawaabta muuji

    \(\frac{-10}{3}\)

  24. \(\underset{x\to 2}{\lim }(\frac{{x}^{2}-5x+6}{x+2})\)

  25. \(\underset{x\to 3}{\lim }(\frac{{x}^{2}-9}{x-3})\)

    Jawaabta muuji

    6

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

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

    Jawaabta muuji

    \(\frac{1}{2}\)

  28. \(\underset{x\to -\frac{7}{2}}{\lim }(\frac{8{x}^{2}+18x-35}{2x+7})\)

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

    Jawaabta muuji

    6

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

  31. \(\underset{x\to 3}{\lim }(\frac{{x}^{2}+2x-3}{x-3})\)

    Jawaabta muuji

    does not exist

  32. \(\underset{h\to 0}{\lim }(\frac{{(3+h)}^{3}-27}{h})\)

  33. \(\underset{h\to 0}{\lim }(\frac{{(2-h)}^{3}-8}{h})\)

    Jawaabta muuji

    \(-12\)

  34. \(\underset{h\to 0}{\lim }(\frac{{(h+3)}^{2}-9}{h})\)

  35. \(\underset{h\to 0}{\lim }(\frac{\sqrt{5-h}-\sqrt{5}}{h})\)

    Jawaabta muuji

    \(-\frac{\sqrt{5}}{10}\)

  36. \(\underset{x\to 0}{\lim }(\frac{\sqrt{3-x}-\sqrt{3}}{x})\)

  37. \(\underset{x\to 9}{\lim }(\frac{{x}^{2}-81}{3-\sqrt{x}})\)

    Jawaabta muuji

    \(-108\)

  38. \(\underset{x\to 1}{\lim }(\frac{\sqrt{x}-{x}^{2}}{1-\sqrt{x}})\)

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

    Jawaabta muuji

    1

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

Ku hay shaqadaada

A xisaab bilaash ah oo ku daray qoraalada waxbarasho kasta, diiwaanka waxa aad dhamaatay, dhibaatooyinka aad xal u tahay meel kaliya, iyo macallinka aad weydiisan kartaa oo ku saabsan bogga. Xisaabta oo keliya waa u furan yahay qof kasta, la diiwaan geliyey ama aan.

Ku soo biir Ku soo gal

Calaamadaha la isticmaalay halkan

Taab calaamad kasta oo loogu talagalay qeexidda buuxda, sawir, iyo waxa xarfaha oo dhan ka mid ah ka dhigan tahay.

Sida loo: Finding Limits: Properties of Limits

  1. Find the limit of a sum, a difference, and a product.
  2. Find the limit of a polynomial.
  3. Find the limit of a power or a root.
  4. Find the limit of a quotient.
  5. Use the properties of limits to break up the polynomial into individual terms.
  6. Find the limits of the individual terms.
  7. Add the limits together.
  8. Alternatively, evaluate the function for

Su'aalaha dadku waydiiyaan

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.

Qaybaha ee boggaan waxaa la beddelay ka OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed iyo dib-oo halkan lagu sharaxay; khaladaadka waa inoo. Sababtoo ah asalka waxaa loo oggolaaday in aan loo isticmaalin ganacsi, fasalkan waa bilaash, waxba ma iibiyo, iyo waxaa lagu wadaagayaa liisan la mid ah.

In ka badan Calculus

LimitsDerivativesIntegralsDefinite integralsTaylor seriesSeries and sumsMaxima and minimaThe chain ruleImplicit differentiationRelated rates and optimisationIntegration techniques: substitution, parts, partial fractionsApplications of integration: area, volume, arc lengthInfinite series and convergence tests