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.

Or, expand in powers of t
$$
\lim_{t->0} \frac{e^2(e^t-1)-t^2 \sin t}{\sin t^2 + \sin t} \\
=
\lim_{t->0} \frac{e^2(t + t^2/2 + \ldots) - (t^3 - t^4/6 + \ldots)}{t^2 - t^6/6 + \ldots + t-t^3/6+\ldots} \\
= e^2
$$
