Evaluating $\lim_{x \to 2} \frac{e^x - e^2 - (x^2 - 4x + 4)\sin (x - 2)}{\sin [(x - 2)^2] + \sin (x-2)}$ I make a mistake somewhere but I cannot find where. The answer is supposed to be $e^2$. I think it can be solved with l'Hopital rule, but that is tedious and error-prone. I was looking for a faster way.
$$\begin{align}
\lim_{x \to 2} \frac{e^x - e^2 - (x^2 - 4x + 4)\sin (x - 2)}{\sin[(x - 2)^2] + \sin (x-2)} &= \lim_{x \to 2} \frac{e^x - e^2 - (x - 2)^2\sin (x - 2)}{\sin[(x - 2)^2] + \sin (x-2)} =\\
&=\left [ t = x - 2, x \to 2 \implies t \to 0 \right ] =\\
&=\lim_{t \to 0} \frac{e^{t + 2} - e^2 - t^2\sin t}{\sin(t^2) + \sin t} =\\
&=\lim_{t \to 0} \frac{e^2 - e^2 - t^2\sin t}{\sin(t^2) + \sin t} =\\
&=\lim_{t \to 0} \frac{- t^2(t + o(t))}{t^2 + o(t^2) + t + o(t)} = \tag{1} \\
&=\lim_{t \to 0} \frac{-t^3}{t} = 0
\end{align}
$$
I think I may be making a mistake in $(1)$ by either simplifying away the two $e$'s or rewriting the expression with the little-o notation.
 A: Setting $x-2=h,$
$$\lim_{x \to 2} \frac{e^x - e^2 - (x^2 - 4x + 4)\sin (x - 2)}{\sin[(x - 2)^2] + \sin (x-2)} $$
$$=e^2\lim_{h\to0}\frac{(e^h-1)}{\sin h^2+\sin h}-\lim_{h\to0}\frac{h^2\sin h}{\sin h^2+\sin h}$$
$$\text{Now,}\lim_{h\to0}\frac{e^h-1}{\sin h^2+\sin h}=\frac{\lim_{h\to0}\frac{(e^h-1)}h}{\lim_{h\to0}\frac{\sin h^2}{h^2}h+\lim_{h\to0}\frac {\sin h}h}=\frac1{1\cdot0+1}$$
$$\text{and }\lim_{h\to0}\frac{h^2\sin h}{\sin h^2+\sin h}=\frac{\lim_{h\to0}h^2\lim_{h\to0}\frac{\sin h}h}{\lim_{h\to0}\frac{\sin h^2}{h^2}\lim_{h\to0} h+\lim_{h\to0}\frac{\sin h}h}=\frac{0\cdot1}{1\cdot0+1}=0$$
as $\displaystyle h\to0\implies\sin h\to0\implies\sin h\ne0$ and $\displaystyle\lim_{h\to0}\frac{e^h-1}h=\lim_{h\to0}\frac{\sin h}h=1 $
A: Although it's tempting you cannot lose this $t$.
$$\begin{align}
&=\lim_{t \to 0} \frac{e^{\color{red}{t} + 2} - e^2 - t^2\sin t}{\sin(t^2) + \sin t} =\\
&=\lim_{t \to 0} \frac{e^2 - e^2 - t^2\sin t}{\sin(t^2) + \sin t} =\end{align}$$
In general, you do not have the freedom of evaluating specific parts of your limit at once. The term with $e^{t+2}$  includes a denominator that goes to $0$. This is the source of your problem.
A: Or, expand in powers of t
$$
\lim_{t->0} \frac{e^2(e^t-1)-t^2 \sin t}{\sin t^2 + \sin t} \\
=
\lim_{t->0} \frac{e^2(t + t^2/2 + \ldots) - (t^3 - t^4/6 + \ldots)}{t^2 - t^6/6 + \ldots + t-t^3/6+\ldots} \\
= e^2
$$
