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.

## Limits at Infinity Rules

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 }$$\frac{1}{x^{t}} = 0 where q is an odd integer. Limit at positive infinity of a polynomial = \infty if the leading coefficient is positive and -\infty if the leading coefficient is negative. ## Limits at Infinity Examples Limits at infinities are found to determine the horizontal asymptotes. ### Solved Examples Question 1: Find \lim_{x\rightarrow \infty }$$\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 }$$\frac{2-\frac{1}{x}+\frac{1}{x^{2}}}{4-\frac{3}{x}-\frac{1}{x^{2}}} = \frac{\lim_{x\rightarrow \infty }(2-\frac{1}{x}+\frac{1}{x^{2}})}{\lim_{x\rightarrow \infty }(4-\frac{3}{x}-\frac{1}{x^{2}})} Quotient Rule = \frac{2}{4} = \frac{1}{2} As \frac{1}{x} and higher powers approach zero as x \to \infty The graph of the rational function also confirms this and it can be seen that y = \frac{1}{2} is the horizontal asymptote to the graph. Question 2: Find \lim_{x\rightarrow \infty }tan^{-1}x and \lim_{x\rightarrow -\infty }tan^{-1}x Solution: \lim_{x\rightarrow \infty }tan^{-1}x = tan-1 (∞) = \frac{\pi }{2} \lim_{x\rightarrow -\infty }tan^{-1}x = tan-1 (-∞) = -\frac{\pi }{2} The graph of \tan^{-1} (x) is shown below with asymptotes y = \frac{\pi }{2} and y = -\frac{\pi }{2} Question 3: Show that \lim_{x\rightarrow \infty }\sqrt{x^{2}+1}-x = 0 Solution: This limit problem is solved using the technique applied for other limit problems. The expression is multiplied and divided by the conjugate expression. \lim_{x\rightarrow \infty }(\sqrt{x^{2}+1}-x).$$\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.