Limits at infinity describe the behavior of a function f(x) as x increases without bounds or decreases without bounds. The notation $\lim_{x\rightarrow \infty }f(x)$ refers to the limit of the function as x is positive and increasingly large. The corresponding notation when x is negative and increasingly large is $\lim_{x\rightarrow -\infty }f(x)$.

There are three possibilities at such situations:

  1. The function can increase without bounds and approach $\infty$.
  2. The function can decrease without bounds and approach $-\infty$.
  3. The function can approach a finite value L as limit at infinity.

In the last case we say the limit at infinity exists. When the limit of a function at infinity is a finite value L, then graph of the function has y = L as the horizontal asymptote.

The following rules similar to general rules for Limits hold for limits at infinity.
Let $\lim_{x\rightarrow \pm \infty }f(x)$ = $L$ and $\lim_{x\rightarrow \pm \infty }g(x)$ = $M$. Then,
  1. $\lim_{x\rightarrow \pm \infty }(f(x)+g(x))$ = $L + M$ Sum Rule
  2. $\lim_{x\rightarrow \pm \infty }(f(x)-g(x))$ = $L - M$ Difference Rule
  3. $\lim_{x\rightarrow \pm \infty }(f(x).g(x))$ = $L \times M$ Product Rule
  4. $\lim_{x\rightarrow \pm \infty }(k.f(x))$ = $k \times L$ Constant Rule
  5. $\lim_{x\rightarrow \pm \infty }$$\frac{f(x)}{g(x)}$ = $\frac{L}{M}$ $M \neq 0$ Quotient Rule
  6. $\lim_{x\rightarrow \pm \infty }(f(x))^{\frac{p}{q}}=L^{\frac{p}{q}}$ $q \neq 0$ Power Rule

Other than these general rules, there are two standard limits at infinity used in solving limit problems.
For a positive rational number $t$ of the form $\frac{p}{q}$

$\lim_{x\rightarrow \infty }$$\frac{1}{x^{t}}$ = $0$

and

$\lim_{x\rightarrow -\infty }$$\frac{1}{x^{t}}$ = $0$ where q is an odd integer.

Limit at positive infinity of a polynomial = $\infty$ if the leading coefficient is positive and $-\infty$ if the leading coefficient is negative.

Limits at infinities are found to determine the horizontal asymptotes.

Solved Examples

Question 1: Find $\lim_{x\rightarrow \infty }$$\frac{2x^{2}-x+1}{4x^2-3x-1}$
Solution:
 
$\lim_{x\rightarrow \infty }$$\frac{2x^{2}-x+1}{4x^2-3x-1}$ = $\lim_{x\rightarrow \infty }$$\frac{2x^{2}-x+1}{4x^{2}-3x-1}$.$\frac{(\frac{1}{x^{2}})}{(\frac{1}{x^{2}})}$    Multiply and divide the numerator and denominator by $\frac{1}{x^{2}}$
    
= $\lim_{x\rightarrow \infty }$$\frac{2-\frac{1}{x}+\frac{1}{x^{2}}}{4-\frac{3}{x}-\frac{1}{x^{2}}}$

     = $\frac{\lim_{x\rightarrow \infty }(2-\frac{1}{x}+\frac{1}{x^{2}})}{\lim_{x\rightarrow \infty }(4-\frac{3}{x}-\frac{1}{x^{2}})}$     Quotient Rule
    
 = $\frac{2}{4}$ = $\frac{1}{2}$                     As $\frac{1}{x}$ and higher powers approach zero as $x \to \infty$

The graph of the rational function also confirms this and it can be seen that y = $\frac{1}{2}$ is the horizontal asymptote to the graph.

Limits at Infinity


 

Question 2: Find $\lim_{x\rightarrow \infty }tan^{-1}x$  and $\lim_{x\rightarrow -\infty }tan^{-1}x$
Solution:
 
$\lim_{x\rightarrow \infty }tan^{-1}x$  = tan-1 (∞) = $\frac{\pi }{2}$

            $\lim_{x\rightarrow -\infty }tan^{-1}x$ = tan-1 (-∞) = $-\frac{\pi }{2}$

   The graph of $\tan^{-1} (x)$ is shown below with asymptotes y = $\frac{\pi }{2}$  and y = $-\frac{\pi }{2}$
 
   Limits at Infinity Example

 

Question 3: Show that $\lim_{x\rightarrow \infty }\sqrt{x^{2}+1}-x$ = 0
Solution:
 
This limit problem is solved using the technique applied for other limit problems. The expression is multiplied and divided by the conjugate expression.

    $\lim_{x\rightarrow \infty }(\sqrt{x^{2}+1}-x).$$\frac{\sqrt{x^{2}+1}+x}{\sqrt{x^{2}+1}+x}$

    = $\lim_{x\rightarrow \infty }$$\frac{(x^{2}+1)-x^{2}}{\sqrt{x^{2}+1}+x}$ = $\lim_{x\rightarrow \infty }$$\frac{1}{\sqrt{x^{2}+1}+x}$

    = $\lim_{x\rightarrow \infty }$$\frac{\frac{1}{x}}{\frac{\sqrt{x^{2}+1}+x}{x}}$      Divide the numerator and denominator by x.

    = $\lim_{x\rightarrow \infty }$$\frac{\frac{1}{x}}{\sqrt{1+\frac{1}{x^{2}}}+1}$     Terms in the denominator divided separately.

    = $\frac{0}{\sqrt{1+0}+1}$ = $\frac{0}{2}$ = 0.