Locating and classifying the vertical asymptote using L'Hopital's Rule

Substituting $x=\pi$ gives $\sin\pi = 0$ and $1+\cos\pi = 1-1=0$, confirming the form $0/0$, so L'Hopital's Rule applies. Differentiating numerator and denominator separately gives $\frac{d}{dx}\sin x = \cos x$ and $\frac{d}{dx}(1+\cos x) = -\sin x$, so the limit becomes $\displaystyle\lim_{x\to \pi} \frac{\cos x}{-\sin x}$.

As $x \to \pi$, $\cos x \to -1$ but $\sin x \to 0$, so this new ratio $\frac{\cos x}{-\sin x}$ is no longer of an indeterminate form — it actually blows up, since the denominator $-\sin x \to 0$ while the numerator $\to -1 \neq 0$. This tells us the original limit does not exist (it is infinite), confirming that $x=\pi$ is a vertical asymptote rather than a removable point.

To determine the one-sided behavior, note that near $x=\pi$, $\sin x$ is small and $1+\cos x \to 0^+$ from both sides (since $\cos x \le 1$ always, $1+\cos x \ge 0$). For $x$ slightly less than $\pi$, $\sin x > 0$, so $u(x) \to +\infty$; for $x$ slightly greater than $\pi$, $\sin x < 0$, so $u(x) \to -\infty$. Thus $x=\pi$ is a vertical asymptote with $u(x)\to +\infty$ as $x\to\pi^-$ and $u(x)\to -\infty$ as $x\to\pi^+$.