# Limits – I

## Limits – I

Limits of The Form $$\underset{x\to a}{\mathop{\lim }}\,{{\left( f(x) \right)}^{g(x)}}$$.

1. Form 0⁰, ∞⁰.

Let  $$L=\underset{x\to a}{\mathop{\lim }}\,{{\left( f(x) \right)}^{g(x)}}$$ then

$${{\log }_{e}}L={{\log }_{e}}\left[ \underset{x\to a}{\mathop{\lim }}\,{{\{f(x)\}}^{g(x)}} \right]$$.

$${{\log }_{e}}L=\underset{x\to a}{\mathop{\lim }}\,\left[ {{\log }_{e}}{{\{f(x)\}}^{g(x)}} \right]$$.

$${{\log }_{e}}L=\underset{x\to a}{\mathop{\lim }}\,g(x){{\log }_{e}}\left[ f(x) \right]$$.

$$L={{e}^{\underset{x\to a}{\mathop{\lim }}\,g(x)\left[ {{\log }_{e}}(f(x) \right])}}$$.

Example 1: Evaluate $$\underset{x\to \infty }{\mathop{\lim }}\,{{x}^{1/x}}$$.

Solution: Given that $$\underset{x\to \infty }{\mathop{\lim }}\,{{x}^{1/x}}={{e}^{\log \underset{x\to \infty }{\mathop{\lim }}\,{{x}^{\frac{1}{x}}}}}$$.

$$={{e}^{\underset{x\to \infty }{\mathop{\lim }}\,\log {{x}^{\frac{1}{x}}}}}$$.

$$={{e}^{\underset{x\to \infty }{\mathop{\lim }}\,\frac{1}{x}\log x}}$$.

(since x increases faster than loge x when x → ∞)

= e⁰

= 0

Example 2: Evaluate $$\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,{{(\cos x)}^{\cos x}}$$.

Solution: Given that $$\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,{{(\cos x)}^{\cos x}}$$.

Let $$y=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,{{(\cos x)}^{\cos x}}$$.

$$\log y=\log \underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,{{(\cos x)}^{\cos x}}$$.

$$\log y=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,\log {{(\cos x)}^{\cos x}}$$.

$$\log y=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,\cos x.\log (\cos x)$$.

$$\log y=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,\frac{1}{\sec x}.\log (\cos x)$$.

Using L – Hospital’s rule

$$=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,\frac{\frac{-\sin x}{\cos x}}{\sec x.\tan x}$$.

$$=\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,(-\cos x)$$.

= 0

logy = 0

y = e⁰

y = 1