Question

Evaluate:
$\rm \displaystyle \lim_{x \to 0}\ \bigg(\frac{1-x}{1+x}\bigg)^{\frac{2}{x}}$

Answer 1

we have to evaluate,$\rm \displaystyle \lim_{x \to 0}\ \bigg(\frac{1-x}{1+x}\bigg)^{\frac{2}{x}}$

we know, if we put, x = 0 you will get 1^∞ is the form of limit.

we know, if $\displaystyle\lim_{x\to 0}f(x)^{g(x)}$ is in the form of 1^∞

then, $\displaystyle\lim_{x\to 0}f(x)^{g(x)}=\displaystyle\lim_{x\to 0}e^{g(x)[f(x)-1]}$

so, $\rm \displaystyle \lim_{x \to 0}\ \bigg(\frac{1-x}{1+x}\bigg)^{\frac{2}{x}}$ = $\displaystyle\lim_{x\to 0}e^{\frac{2}{x}\left(\frac{1-x}{1+x}-1\right)}$

= $e^{\displaystyle\lim_{x\to 0}\frac{2}{x}\times\frac{1-x-1-x}{1+x}}$

= $e^{\displaystyle\lim_{x\to 0}\frac{2}{x}\times\frac{-2x}{1+x}}$

= $e^{\displaystyle\lim_{x\to 0}\frac{-4}{1+x}}$

= e^-4 = 1/e⁴ [Ans]

Answer 2

Answer:

$\frac{1}{e^{4} }$

Step-by-step explanation:

Given:

$\lim_{x \to 0} (\frac{1-x}{1+x} )^{\frac{2}{x} }$

Put x=0 in the  above question we get:

$\lim_{x \to 0}( \frac{1-0}{1+0}  )^{\frac{2}{0} }$

=$1^{\infty}$ Form

Now we are using following  formula:

$\lim_{x \to 0} f(x)^{g(x)}$

If in the form of $1^{\infty}$

Then:

$\lim_{x \to 0} e^{g(x)[f(x)-1]}$

$\lim_{x \to 0} e^{\frac{2}{x}(\frac{1-x}{1+x}-1)  }$

$e^{ \lim_{x \to 0} \frac{2}{x} (\frac{1-x-1-x}{1+x})  }$

$e^{ \lim_{x \to 0} \frac{2}{x}( \frac{-2x}{1+x})  }$

$e^{ \lim_{x \to 0} \frac{-4}{1+x} }$

$e^{-4}$

$\frac{1}{e^{4} }$