Question

limit x tends to 0 e
${ e}^{ - 2x} - 1 \div x$
​

Answer 1

Step-by-step explanation:

given :

• limit x tends to 0 e
• ${ e}^{ - 2x} - 1 \div x$

to find :

• tex] { e}^{ - 2x} - 1 \div x[/tex]

solution :

• hence, the answer is 0

Answer 2

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\sf \frac{ {e}^{ - 2x} - 1 }{x}$

If we substitute directly x = 0, we get

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

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

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

which is indeterminant form.

So, Consider again

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\sf \frac{ {e}^{ - 2x} - 1 }{x}$

can be rewritten as

$\rm \: = \: \displaystyle\lim_{x \to 0}\sf \frac{ {e}^{ - 2x} - 1 }{ - 2x} \times ( - 2)$

$\rm \: = \: - 2 \: \displaystyle\lim_{x \to 0}\sf \frac{ {e}^{ - 2x} - 1 }{ - 2x}$

We know,

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

So, using this identity, we get

$\rm \: = \: - 2 \times 1$

$\rm \: = \: - 2$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0}\sf \frac{ {e}^{ - 2x} - 1 }{x} \: = \: - \: 2 \: }}$

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Additional Information :-

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

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

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

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

$\boxed{ \tt{ \: \displaystyle\lim_{x \to a}\sf \: \frac{ {x}^{n} - {a}^{n} }{x - a} = {na}^{n - 1} \: }}$