$$f(x) = \lim_{n \to \infty}\frac{x^2 e^{nx} + \cos(x)}{e^{nx}+1}, x \in \mathbb{R}.$$
We need to check whether the function $f(x)$ is continuous.
$$g(x) = \frac{2xe^{nx}}{ne^{nx-1}} + x^3 - \frac{\sin(x)}{ne^{nx-1}}$$ (using L' Hopitals).
$$\lim_{n \to \infty}\left(\frac{2xe^{nx}}{ne^{nx-1}} + x^3 - \frac{\sin(x)}{ne^{nx-1}}\right) = x^3$$
I used the L'Hospital and found that the limit of the function $f(x)$ is $x^3$.
How do I proceed after this?
If $x<0$ then $e^{nx}\to 0$ as $n\to\infty$ so $f(x)=\frac {x^2\cdot 0+\cos x}{0+1}=\cos x. $
If $x=0$ then $\frac {x^2e^{nx}+\cos x}{e^{nx}+1}=\frac {\cos 0}{2}=1/2$ for every $n$ so $f(0)=1/2.$
If $x>0$ then $e^{nx}+1\to \infty$ as $n\to \infty$, and $\frac {|\cos x|}{|e^{nx}+1|}\le \frac {1}{e^{nx}+1},$ so $\frac {\cos x}{e^{nx}+1}\to 0.$ And $\frac {x^2e^{nx}}{e^{nx}+1}=x^2-x^2\frac {1}{e^{nx}+1}\to x^2.$ So $f(x^2)=x^2.$