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Evaluating \mathrm{\lim _{x \rightarrow \infty} \frac{\sum_{r=1}^{x} r e^{\frac{r}{x}}}{x^{2}}}

Option: 1

1


Option: 2

0


Option: 3

2


Option: 4

-3


Answers (1)

best_answer

\mathrm{\lim _{x \rightarrow \infty}\left(\frac{e^{\frac{1}{x}}+2 e^{\frac{2}{x}}+3 e^{\frac{3}{x}}+4 e^{\frac{4}{x}}+\ldots \ldots+x e^{\frac{x}{x}}}{x^{2}}\right)}

By subsituting \mathrm{e^{\frac{1}{x}}=t}
\mathrm{\lim _{x \rightarrow \infty}\left(\frac{t+2 t^{2}+3 t^{3}+4 t^{4}+\ldots \ldots+x t^{x}}{x^{2}}\right)}

to keep things simple let us first calculatethe value of \mathrm{t+2 t^{2}+3 t^{3}+4 t^{4}+\ldots \ldots+x t^{x}}\quad \ldots3
\mathrm{S=t+2 t^{2}+3 t^{3}+4 t^{4}+\ldots \ldots+x t^{x}}\quad \ldots 1

dividing both sides by \mathrm{t}
\mathrm{\frac{S}{t}=1+2 t^{1}+3 t^{2}+4 t^{3}+\ldots \ldots+x t^{x-1}}\quad \ldots 2

Subtarcting 1 from 2
\mathrm{S\left(\frac{1-t}{t}\right)=1+t^{1}+t^{2}+t^{3}+\ldots .+t^{x-1}-x t^{x}}

Using the sum of geometric progression
\mathrm{S\left(\frac{1-t}{t}\right)=\frac{t^{x}-1}{t-1}-x t^{x}}

Using the sum of geometric progression
\mathrm{S\left(\frac{1-t}{t}\right)=\frac{t^{x}-1}{t-1}-x t^{x} }
\mathrm{S\left(\frac{1-t}{t}\right)=\frac{t^{x}-1-x t^{x+1}+x t^{x}}{t-1} }
\mathrm{S=\frac{t^{x+1}-t-x t^{x+2}+x t^{x+1}}{-(t-1)^{2}} }

Substituting back \mathrm{t=e^{\frac{1}{x}}}
\mathrm{S=\frac{-e^{1+\frac{1}{x}}+e^{\frac{1}{x}}+x e^{1+\frac{2}{x}}-x e^{1+\frac{1}{x}}}{\left(e^{\frac{1}{x}}-1\right)^{2}}}

Putting \mathrm{\mathrm{S}} back in the equation 3
\mathrm{\lim _{x \rightarrow \infty} \frac{-e^{1+\frac{1}{x}}+e^{\frac{1}{x}}+x e^{1+\frac{2}{x}}-x e^{1+\frac{1}{x}}}{x^{2}\left(e^{\frac{1}{x}}-1\right)^{2}}}
\mathrm{\lim _{x \rightarrow \infty} \frac{-e^{1+\frac{1}{x}}+e^{\frac{1}{x}}+\frac{x e^{1+\frac{1}{x}}\left(e^{\frac{1}{x}}-1\right) x}{x}}{x^{2} \frac{\left(e^{\frac{1}{x}}-1\right)^{2}}{x^{2}} x^{2}}}

By applying the standard limit
\mathrm{\lim _{x \rightarrow 0} \frac{a^{x}-1}{x}=\log _{e} a}
\mathrm{\frac{-e+1+e}{1}}

Posted by

sudhir kumar

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