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Find  \mathrm{\lim _{x \rightarrow \infty} e^{2 x \cdot \ln \frac{x+1}{x-2}}}

Option: 1

\mathrm{e^{6}}


Option: 2

\mathrm{e^{4}}


Option: 3

\mathrm{e^{2}}


Option: 4

None


Answers (1)

best_answer

\mathrm{\lim _{x \rightarrow \infty} e^{2 x \ln \frac{x+1}{x-2}}=\lim _{x \rightarrow \infty} e^{\ln \left(\frac{x+1}{x-2}\right)^{2 x}}}
                            \mathrm{=\lim _{x \rightarrow \infty}\left(\frac{x+1}{x-2}\right)^{2 x}}
                            \mathrm{=\lim _{x \rightarrow \infty}\left(\frac{(x-2)+3}{x-2}\right)^{2 x}}
                            \mathrm{=\lim _{x \rightarrow \infty}\left(1+\frac{3}{x-2}\right)^{2 x} }
                           \mathrm{=\lim _{x \rightarrow \infty}\left[\left(1+\frac{3}{x-2}\right)^{(x-2)+2}\right]^{2} }
                          \mathrm{ =\lim _{x \rightarrow \infty}\left[\left(1+\frac{3}{x-2}\right)^{(x-2)}\left(1+\frac{3}{x-2}\right)^{2}\right]^{2} }                                       
                        \mathrm{ =\lim _{x \rightarrow \infty}\left[\left(1+\frac{3}{x-2}\right)^{(x-2)}\right]^{2}\left[\left(1+\frac{3}{x-2}\right)^{2}\right]^{2} }     
                      \mathrm{ =\lim _{x \rightarrow \infty}\left[\left(1+\frac{3}{x-2}\right)^{(x-2)}\right]^{2} \lim _{x \rightarrow \infty}\left(1+\frac{3}{x-2}\right)^{4} }
                    \mathrm{ =\left[\lim _{x \rightarrow \infty}\left(1+\frac{3}{x-2}\right)^{(x-2)}\right]^{2} \lim _{x \rightarrow \infty}\left(1+\frac{3}{x-2}\right)^{4} }
                   \mathrm{ =\left[\lim _{(x-2) \rightarrow \infty}\left(1+\frac{3}{x-2}\right)^{(x-2)}\right]^{2} \lim _{x \rightarrow \infty}\left(1+\frac{3}{x-2}\right)^{4} }
                  \mathrm{ =\left(e^{3}\right)^{2} \cdot 1}
                  \mathrm{ =e^{6}}

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Rishabh

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