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  • Evaluating: <img alt="\mathrm{ \lim _{x \rightarrow 0}\left(\frac{x e^x-2+2 \cos x-x}{(\sin x)^2 \tan (2 x)+x e^{-\frac{1}{|x|}}}\right) }" src="https://entrancecorner.oncodecogs.com/gif.lat

Evaluating:
\mathrm{ \lim _{x \rightarrow 0}\left(\frac{x e^x-2+2 \cos x-x}{(\sin x)^2 \tan (2 x)+x e^{-\frac{1}{|x|}}}\right) }

Option: 1

0


Option: 2

1


Option: 3

\frac{1}{2}


Option: 4

1 / 4


Answers (1)

best_answer

Its best to use Taylor expansions here. We have

                                \mathrm{e^x=1+x+\frac{x^2}{2}+o\left(x^2\right), \cos x=1-\frac{x^2}{2}+o\left(x^3\right)}

and

                                  \mathrm{\sin x=x+o(x), \tan 2 x=2 x+o(x)}

Also note that

                                                           \mathrm{e^{-1 /|x|}=o\left(x^n\right)}

for all positive integers n (this fact is the key to solving this problem and requires some effort to prove, but it can be taken as one of the standard limits). It thus follows that the numerator is equal to

                              \mathrm{x+x^2+\frac{x^3}{2}-2+2-x^2-x+o\left(x^3\right)=\frac{x^3}{2}+o\left(x^3\right)}

and the denominator is equal to

                                      \mathrm{x^2 \cdot 2 x+o\left(x^3\right)=2 x^3+o\left(x^3\right)}

and therefore the ratio is equal to

                                          \mathrm{\frac{\frac{x^3}{2}+o\left(x^3\right)}{2 x^3+o\left(x^3\right)}=\frac{1+o(1)}{4+o(1)}}

and thus it tends to \mathrm{1 / 4 \text { as } x \rightarrow 0 \text {. }}

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