Question

Evaluate the limit . ​

Answer 1

Given Question :-

Evaluate the following

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \displaystyle\lim_{x \to 0} \frac{ {a}^{x} - {b}^{x} }{x \sqrt{1 - {x}^{2} } }$

$\red{\large\underline{\sf{Solution-}}}$

Given expression is

$\rm :\longmapsto\: \: \displaystyle\lim_{x \to 0} \frac{ {a}^{x} - {b}^{x} }{x \sqrt{1 - {x}^{2} } }$

If we substitute directly x = 0, we get

$\rm \: = \:\dfrac{ {a}^{0} - {b}^{0} }{0 \times \sqrt{1 - 0} }$

$\rm \: = \:\dfrac{1 - 1}{0}$

$\rm \: = \:\dfrac{0}{0}$

which is indeterminant form.

So, above expression

$\rm :\longmapsto\: \: \displaystyle\lim_{x \to 0} \frac{ {a}^{x} - {b}^{x} }{x \sqrt{1 - {x}^{2} } }$

can be rewritten as

$\rm \: = \:\displaystyle\lim_{x \to 0} \: \frac{ {a}^{x} - {b}^{x} }{x} \times \displaystyle\lim_{x \to 0} \frac{1}{ \sqrt{1 - {x}^{2} } }$

$\rm \: = \:\displaystyle\lim_{x \to 0} \: \frac{ {a}^{x} - {b}^{x} - 1 + 1 }{x} \times \frac{1}{ \sqrt{1 - {0}^{2} } }$

$\rm \: = \:\displaystyle\lim_{x \to 0} \: \frac{ [{a}^{x} - 1] - [ {b}^{x} - 1]}{x} \times \frac{1}{ \sqrt{1} }$

$\rm \: = \:\displaystyle\lim_{x \to 0} \frac{ {a}^{x} - 1}{x} - \displaystyle\lim_{x \to 0} \frac{ {b}^{x} - 1}{x}$

$\rm \: = \:loga - logb$

$\rm \: = \:log\bigg | \dfrac{a}{b} \bigg|$

More to know,

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{sinx}{x} = 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{tanx}{x} = 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{log(1 + x)}{x} = 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{ {e}^{x} - 1 }{x} = 1 \: }}$

$\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{ {a}^{x} - 1 }{x} = loga \: }}$