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Wize AP Calculus (BC) Textbook > Limits & Continuity

Limits at Infinity

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Limits at Infinity

Here we would like to know what happens to a function if xx becomes arbitrarily large (either positively or negatively).


Limits at (Positive) Infinity

If a function f(x)f\left(x\right) approaches LL as xx grows arbitrarily large, then
lim⁡x→∞f(x)=L \displaystyle \boxed{\lim_{x \to \infty} f(x)=L}

Limits at (Negative) Infinity

If a function f(x)f(x) approaches LL as xx grows arbitrarily small, then
lim⁡x→−∞f(x)=L \displaystyle \boxed{\lim_{x \to -\infty} f(x)=L}


Wize Tip
lim⁡x→±∞1xN=0\displaystyle \lim_{x\rightarrow \pm\infin}\frac{1}{x^N}=0 , for any N>0N>0

Since 1 divided by something "very large" is very close to 0.



Wize Tip
If you are dealing with powers of in a limit, it might help to factor out the largest power of in the numerator and denominator (if it’s a fraction)




Horizontal Asymptotes

A function f(x)f\left(x\right) has a horizontal asymptote with equation y=Ly=L if at least one of these statements is true:

lim⁡x→∞f(x)=L    or    lim⁡x→−∞f(x)=L\displaystyle \boxed{\lim_{x \to \infty} f(x)=L\;\; \text{or} \;\; \displaystyle \lim_{x \to -\infty} f(x)=L}



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Example: Limits at Infinity

For the following graph, find:

i) lim⁡x→∞f(x)\displaystyle\lim_{x \to\infty} f(x)

ii)lim⁡x→−∞f(x)\displaystyle\lim_{x \to-\infty} f(x)



Note: that the lines y=±3y=\pm3 are horizontal asymptotes of f(x)f(x)

i) lim⁡x→∞f(x)=3\displaystyle\lim_{x \to\infty} f(x)=3

ii) lim⁡x→−∞f(x)=−3\displaystyle\lim_{x \to-\infty} f(x)=-3
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Example: Limits at Infinity

Find the following limit

lim⁡x→∞x3+x−12x3+3x2+2\displaystyle\lim_{x\rightarrow\infty}\frac{x^3+x-1}{2x^3+3x^2+2}


lim⁡x→∞x3+x−12x3+3x2+2∗1x31x3=lim⁡x→∞1+1x2−1x32+3x+2x3=1+0+02+0+0=12\displaystyle\lim_{x\rightarrow\infty}\frac{x^3+x-1}{2x^3+3x^2+2}*\frac{\frac{1}{x^3}}{\frac{1}{x^3}} \\ \text{} \\=\lim_{x\rightarrow\infty}\frac{1+\frac{1}{x^2}-\frac{1}{x^3}}{2+\frac{3}{x}+\frac{2}{x^3}}\\ \text{} \\=\frac{1+0+0}{2+0+0}\\ \text{} \\=\boxed{\frac{1}{2}}









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Example: Finding Asymptotes with Limits


Find the horizontal and vertical asymptotes of f(x)=1x+5f(x) = \frac{1}{x+5} by using infinite limits and limits at infinity.


Vertical Asymptote(s)
  • lim⁡x→−5+1x+5→1−5+5→10+→∞\displaystyle\lim_{x\rightarrow -5^+}\frac{1}{x+5} \rightarrow \frac{1}{-5+5} \rightarrow \frac{1}{0^+} \rightarrow \infty (here 0+0^+denotes approaching 0 from positive values of x)
  • lim⁡x→−5−1x+5→1−5+5→10−→−∞\displaystyle\lim_{x\rightarrow -5^-}\frac{1}{x+5} \rightarrow \frac{1}{-5+5} \rightarrow \frac{1}{0^-} \rightarrow -\infty(here 0−0^- denotes approaching 0 from negative values of x)
Therefore, there is a vertical asymptote at x=−5x=-5.

Note: this is the value for xxwhich makes the denominator 0.

Horizontal Asymptote(s)
  • lim⁡x→∞1x+5→1∞+5→0\displaystyle\lim_{x\rightarrow \infty}\frac{1}{x+5} \rightarrow \frac{1}{\infty+5} \rightarrow 0
  • lim⁡x→−∞1x+5→1−∞+5→0\displaystyle\lim_{x\rightarrow -\infty}\frac{1}{x+5} \rightarrow \frac{1}{-\infty+5} \rightarrow 0
Therefore, there is a horizontal asymptote at y=0y=0.

Note: we also know this since the degree of the polynomial in the denominator is larger than the degree of the polynomial in the numerator.
lim⁡x→∞9x6−xx3+1=\displaystyle \lim_{x \to \infty} \frac{\sqrt{9x^6-x}}{x^3+1}=
Evaluate lim⁡x→−∞3x6−x+12x3\displaystyle\lim_{x\to-\infty}\frac{\sqrt{3x^6-x+1}}{2x^3}

*Be careful with the negative sign in −∞-\infty! Check out the hint for more info.