Math, asked by Anonymous, 30 days ago

How to solve the limit x^n/e^x where x tends to infinity?

Answers

Answered by mathdude500
9

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

Given expression is

\rm \: \displaystyle\lim_{x \to  \infty }\rm  \frac{ {x}^{n} }{ {e}^{x} }  \\

On applying L Hospital Rule, we get

\rm \:  =  \: \displaystyle\lim_{x \to  \infty }\rm  \frac{ \dfrac{d}{dx}{x}^{n} }{ \dfrac{d}{dx}{e}^{x} }  \\

\rm \: =  \:\displaystyle\lim_{x \to  \infty }\rm  \frac{n {x}^{n - 1} }{ {e}^{x} }  \\

Again, on applying L Hospital Rule, we get

\rm \: =  \:\displaystyle\lim_{x \to  \infty }\rm  \frac{n(n - 1) {x}^{n - 2} }{ {e}^{x} }  \\

Again, on applying L Hospital Rule, we get

\rm \: =  \:\displaystyle\lim_{x \to  \infty }\rm  \frac{n(n - 1)(n - 2) {x}^{n - 3} }{ {e}^{x} }  \\

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and proceeding like this,

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On applying L Hospital Rule, we get

\rm \: =  \:\displaystyle\lim_{x \to  \infty }\rm  \frac{n(n - 1)(n - 2)(n - 3)...1}{ {e}^{x} }  \\

\rm \: =  \:\displaystyle\lim_{x \to  \infty }\rm  \frac{n!}{ {e}^{x} }  \\

\rm \: =  \:\displaystyle\lim_{x \to  \infty }\rm n! \:  {e}^{ - x}  \\

\rm \: =  \:n! \times  {e}^{ -  \infty }  \\

\rm \: =  \:n! \times 0 \\

\rm \: =  \:0 \\

Hence,

\rm\implies \: \:\boxed{\sf{  \:  \: \rm \: \displaystyle\lim_{x \to  \infty }\rm  \frac{ {x}^{n} }{ {e}^{x} }  \:  =  \: 0 \:  \: \:  }} \\

\rule{190pt}{2pt}

Additional Information :-

\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\  \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf  -  \: sinx \\ \\ \sf tanx & \sf  {sec}^{2}x \\ \\ \sf cotx & \sf  -  {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf  -  \: cosecx \: cotx\\ \\ \sf  \sqrt{x}  & \sf  \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf  {e}^{x}  & \sf  {e}^{x}  \end{array}} \\ \end{gathered}


amansharma264: Excellent
mathdude500: Thank you so much
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