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Answer 1

Step-by-step explanation:

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Answer 2

Given Question

Prove that,

$\displaystyle\lim_{x \to \infty }\rm \bigg[1 + {\dfrac{1}{x} }\bigg]^{x} = e$

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

Consider,

$\rm :\longmapsto\:\displaystyle\lim_{x \to \infty }\rm \bigg[1 + {\dfrac{1}{x} }\bigg]^{x}$

Let assume that

$\rm :\longmapsto\:y = \displaystyle\lim_{x \to \infty }\rm \bigg[1 + {\dfrac{1}{x} }\bigg]^{x}$

Taking log on both sides, we get

$\rm :\longmapsto\: log_{e}(y) = \displaystyle\lim_{x \to \infty }\rm log_{e} \bigg[1 + {\dfrac{1}{x} }\bigg]^{x}$

We know,

$\red{\rm :\longmapsto\:\boxed{\tt{ log {x}^{y} = y \: logx}}}$

So, using this, we get

$\rm :\longmapsto\: log_{e}(y) = \displaystyle\lim_{x \to \infty }\rm xlog_{e} \bigg[1 + {\dfrac{1}{x} }\bigg]$

We know,

$\rm :\longmapsto\:\boxed{\tt{ log_{e}(1 + x) = x - \dfrac{ {x}^{2} }{2} + \dfrac{ {x}^{3} }{3} + - - - - }}$

So, using this, we get

$\rm :\longmapsto\: log_{e}(y) = \displaystyle\lim_{x \to \infty }\rm x \bigg[\dfrac{1}{x} - {\dfrac{1}{ {2x}^{2} } + \dfrac{1}{ {3x}^{3}} - \dfrac{1}{ {4x}^{4} } + - - - }\bigg]$

So, using this ,we get

$\rm :\longmapsto\: log_{e}(y) = \displaystyle\lim_{x \to \infty }\rm \bigg[1 - {\dfrac{1}{ 2x } + \dfrac{1}{ {3x}^{2}} - \dfrac{1}{ {4x}^{3} } + - - - }\bigg]$

$\rm :\longmapsto\: log_{e}(y) = 1$

$\bf\implies \:y = e$

$\bf\implies \:\:\displaystyle\lim_{x \to \infty }\rm \bigg[1 + {\dfrac{1}{x} }\bigg]^{x} = e$

Hence, Proved

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Additional Information

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

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

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

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

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