Step-by-step explanation:
Given Question
Prove that,
[tex]\displaystyle\lim_{x \to \infty }\rm \bigg[1 + {\dfrac{1}{x} }\bigg]^{x} = e[/tex]
[tex] \red{\large\underline{\sf{Solution-}}}[/tex]
Consider,
[tex]\rm :\longmapsto\:\displaystyle\lim_{x \to \infty }\rm \bigg[1 + {\dfrac{1}{x} }\bigg]^{x}[/tex]
Let assume that
[tex]\rm :\longmapsto\:y = \displaystyle\lim_{x \to \infty }\rm \bigg[1 + {\dfrac{1}{x} }\bigg]^{x}[/tex]
Taking log on both sides, we get
[tex]\rm :\longmapsto\: log_{e}(y) = \displaystyle\lim_{x \to \infty }\rm log_{e} \bigg[1 + {\dfrac{1}{x} }\bigg]^{x}[/tex]
We know,
[tex] \red{\rm :\longmapsto\:\boxed{\tt{ log {x}^{y} = y \: logx}}}[/tex]
So, using this, we get
[tex]\rm :\longmapsto\: log_{e}(y) = \displaystyle\lim_{x \to \infty }\rm xlog_{e} \bigg[1 + {\dfrac{1}{x} }\bigg][/tex]
We know,
[tex]\rm :\longmapsto\:\boxed{\tt{ log_{e}(1 + x) = x - \dfrac{ {x}^{2} }{2} + \dfrac{ {x}^{3} }{3} + - - - - }}[/tex]
So, using this, we get
[tex]\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][/tex]
So, using this ,we get
[tex]\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][/tex]
[tex]\rm :\longmapsto\: log_{e}(y) = 1[/tex]
[tex]\bf\implies \:y = e[/tex]
[tex]\bf\implies \:\:\displaystyle\lim_{x \to \infty }\rm \bigg[1 + {\dfrac{1}{x} }\bigg]^{x} = e[/tex]
Hence, Proved
▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬
Additional Information
[tex]\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} = 1}}[/tex]
[tex]\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} = 1}}[/tex]
[tex]\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{log(1 + x)}{x} = 1}}[/tex]
[tex]\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {e}^{x} - 1}{x} = 1}}[/tex]
[tex]\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1}{x} = loga}}[/tex]
