Evaluating $\lim_{x\to3}\frac{e^{-1\over(3-x)^2}+e(4-3\cos(x-3))^{1/5}-e^{\sqrt {4-x}}}{\sqrt {1-\cos(x-3)}}$ First of all, I make the substitution $x-3=t$ which gives me: $$\lim_{t\to0}\frac{e^{-1\over t^2}+e(4-3\cos t)^{1/5}-e^{\sqrt {1-t}}}{\sqrt {1-\cos t}}$$
The term $e^{-1\over t^2}$ goes to $0$ as $t$ approaches $0$ so I ignore it and $e^{\sqrt {1-t}}$ is just $e$, so I have to work on remaining two terms.
The term in the denominator becomes ${t\over \sqrt2}$ as we replace $\cos t$ by $1-{t^2\over2}$. Applying the same to remaining term yields  $e(1+{3t^2\over2})^{1/5}=e(1+{3t^2\over10})$. So we have: $$\lim_{t\to0}{e+{e3t^2\over10}-e\over{t\over\sqrt 2}}$$
Which should equal to $0$, but apparently the correct answer is ${-e\over\sqrt2}$. So where am I making a mistake?
 A: Considering the limit  $$\lim_{x\to3^-}\frac{e^{\frac{-1}{(3-x)^2}}+e(4-3\cos(x-3))^{1/5}-e^{\sqrt {4-x}}}{\sqrt {1-\cos(x-3)}}\to{\small{\begin{bmatrix}
&t=x-3&\\
&t\to0^-&
\end{bmatrix}}}\to\lim_{t\to0^-}\frac{e^{-\frac{1}{t^2}}+e(4-3\cos t)^{1/5}-e^{\sqrt {1-t}}}{\sqrt {1-\cos t}}$$
And, as you mentioned it is enough to consider the limit
$$\lim_{t\to0^-}\frac{e(4-3\cos t)^{1/5}-e^{\sqrt {1-t}}}{\sqrt {1-\cos t}}.$$
Using Taylor Series, we have
$$\begin{aligned}
&(4-3\cos t)^{1/5}\sim 1+\frac{3 t^2}{10}+\cdots \\
&e^{\sqrt{1-t}}\sim e-\frac{e t}{2}+\cdots\\
&\cos(t)\sim1-\frac{t^2}{2}+\cdots
\end{aligned}$$
and substituting (and noting that the asymptotic terms $\to0$ as $t\to0$) we have
$$\lim_{t\to0^-}\frac{e+\frac{3et^2}{10}-e+\frac{e t}{2}}{\frac{\vert t\vert}{\sqrt{2}}}=-\lim_{t\to0^-}\frac{\frac{3et}{10}+\frac{e}{2}}{\frac{1}{\sqrt{2}}}=-\frac{e}{\sqrt{2}}.$$

It is clear, from this last expression, that since there is a $\vert t\vert$ in the denominator the limit will not exist because $x\to3^+\implies t\to0^+\implies \vert t\vert=t$.
