Question

Solve the given question.
$\displaystyle \lim \limits_{x \to0} \left(\frac{(1 + x)^{ \frac{2}{x} } }{ {e}^{2} } \right)^{ \frac{4}{ \sin(x)} }$
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Answer is 1/e⁴​

Answer 1

EXPLANATION.

$\sf \implies \displaystyle \lim_{x \to 0} \bigg[ \dfrac{(1 + x)^{2/x} }{e^{2} } \bigg]^{4/sin(x)}$

As we know that,

First put the value of x = 0 in the equation, we get.

$\sf \implies \displaystyle \lim_{x \to 0} \bigg[ \dfrac{(1 + (0))^{2/0} }{e^{2} } \bigg]^{4/sin(0)} = 1^{\infty}$

We can see that,

It is in the form of $1^{\infty}$ intermediate form.

We can write equation as,

$\sf \implies \displaystyle e^{\lim_{x \to 0}\dfrac{4}{sin(x)}\bigg[ln \bigg( \dfrac{(1 + x)^{2/x} }{e^{2}} \bigg) \bigg] }$

$\sf \implies \displaystyle e^{\lim_{x \to 0}}\dfrac{4}{sin(x)} \bigg[ln(1 + x)^{2/x} - ln(e^{2} ) \bigg]$

$\sf \implies \displaystyle e^{\lim_{x \to 0}}\dfrac{4}{sin(x)} \bigg[ \dfrac{2}{x} ln(1 + x) - 2ln(e) \bigg]$

$\sf \implies \displaystyle e^{\lim_{x \to 0}}\dfrac{4}{sin(x)} \bigg[ \dfrac{2}{x} ln(1 + x) - 2 \bigg]$

$\sf \implies \displaystyle e^{\lim_{x \to 0}}\dfrac{4 \times 2}{sin(x)} \bigg[ \dfrac{ln(1 + x)}{x} - 1 \bigg]$

$\sf \implies \displaystyle e^{\lim_{x \to 0}}\dfrac{8}{sin(x)} \bigg[ \dfrac{ln(1 + x)}{x} - 1 \bigg]$

As we know that,

Expansion of ㏑(1 + x) = x - x²/2 + x³/3 - . . . . .

Using this expansions in the equation, we get.

$\sf \implies \displaystyle e^{\lim_{x \to 0}}\dfrac{8}{sin(x)} \bigg[ \dfrac{x - x^{2} /2 + x^{3} /3 + . . . . .}{x} - 1 \bigg]$

$\sf \implies \displaystyle e^{\lim_{x \to 0}}\dfrac{8}{sin(x)} \bigg[ 1 - \dfrac{x}{2} + \dfrac{x^{2} }{3} + . . . . . - 1 \bigg]$

As we know that,

$\sf \implies \displaystyle {\lim_{x \to 0}}\dfrac{sin(x)}{x} = \dfrac{x}{sin(x)} 1.$

Using this standard limits in the equation, we get.

$\sf \implies \displaystyle e^{\lim_{x \to 0}}\dfrac{8(x)}{sin(x)} \bigg[ \dfrac{ - x/2 + . . . . . }{x} \bigg]$

$\sf \implies \displaystyle e^{8 \times (-1/2)} = e^{-4}$

$\sf \implies \displaystyle \frac{1}{e^{4} }$

$\sf \implies \displaystyle \boxed{\lim_{x \to 0} \bigg[ \dfrac{(1 + x)^{2/x} }{e^{2} } \bigg]^{4/sin(x)} = \dfrac{1}{e^{4} } }$