I have a problem with this limit, I don't know what method to use. I have no idea how to compute it. Is it possible to compute this limit with the McLaurin expansion? Can you explain the method and the steps used? Thanks. (I prefer to avoid to use l'Hospital's rule.)
$$\lim _{x\to 0}\left(\frac{xe^x-2+2\cos x-x}{\left(\sin x\right)^2 \tan\left(2x\right)+xe^{-\frac{1}{\left|x\right|}}}\right)$$
On
The numerator is $x(e^x -1)-2(1-\cos(x))$ and written in series can be divided by $x^3$. The denominator has two parts: First $\sin(x)$ can be devided by $x$ and therefore $\sin(x)^2 \tan(2x)$ by $x^3$, second $x^n e^{-\frac{1}{|x|}}$ with $n\in\mathbb{Z}$ is always $0$ for $x\to 0$ which means you can ignore it.
I think you know now what to do. :-)
On
Use equivalents (usually the first non-zero term in Taylor-McLaurin's expansion).
As $\mathrm e^{-\tfrac1{\lvert x\rvert}}$ does not have a Taylor's expansion in a neighbourhood of $0$, we need a preliminary result:
For all $n$, $\mathrm e^{-\tfrac1{\lvert x\rvert}}=o(x^n)$ as $x\to 0$.
Indeed, set $u=\dfrac 1{\lvert x\rvert}$. The assertion means $$\frac{\mathrm e^{-\tfrac1{\lvert x\rvert}}}{\lvert x\rvert^n}=\frac{u^n}{e^u}\to 0\enspace\text{as}\enspace u\to+\infty.$$ Now $\;\sin^2x\tan 2x=(x^2+o(x^))(2x+o(x))=2x^3+o(x^3)$, so the denominator is $2x^3+o(x^3)\sim_0 2x^3$.
As to the numerator, $$xe^x-2+2\cos x-x=x\Bigl(1+x+\dfrac{x^2}2+o(x^2)\Bigr)-2\Bigl(\dfrac{x^2}2+o(x^3)\Bigr)-x=\dfrac{x^3}2+o(x^3)$$ whence $$\frac{xe^x-2+2\cos x-x}{\sin^2x \tan2x+xe^{-\tfrac1{\lvert x\rvert}}}\sim_0\dfrac{\frac{x^3}2}{2x^3}=\frac14.$$
Its best to use Taylor expansions here. We have $$e^{x} = 1 + x + \frac{x^{2}}{2} + o(x^{2}),\, \cos x = 1 - \frac{x^{2}}{2} + o(x^{3})$$ and $$\sin x = x + o(x), \, \tan 2x = 2x + o(x)$$ Also note that $$e^{-1/|x|} = o(x^{n})$$ 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 $$x + x^{2} + \frac{x^{3}}{2} - 2 + 2 - x^{2} - x + o(x^{3}) = \frac{x^{3}}{2} + o(x^{3})$$ and the denominator is equal to $$x^{2}\cdot 2x + o(x^{3}) = 2x^{3} + o(x^{3}) $$ and therefore the ratio is equal to $$ \dfrac{\dfrac{x^{3}}{2} + o(x^{3})}{2x^{3} + o(x^{3})} = \frac{1 + o(1)}{4 + o(1)}$$ and thus it tends to $1/4$ as $x \to 0$.