MHB 1.4.203 AP calculus practice question on Limits

[karush]

I am posting some AP calculus practice questions on MeWe so thot I would pass them thru here first
The solution is mine...
any typos or suggestions...

$\textbf{Find the Limit of}$
$\displaystyle\lim_{x\to \pi} \dfrac{\cos{x}+\sin{x}+1}{x^2-\pi^2}$
(A) $-\dfrac{1}{2\pi}$
(B) $\dfrac{1}{\pi}$
(C) $1$
(D) DNE
$\textbf{Solution}$
By observation we have $\frac{0}{0}$ so apply L'Hopital's Rule
apply LH'R then plug in $\pi$ and simplify
$$\displaystyle\lim _{x\to \:\pi }
\left(\frac{\cos \left(x\right)-\sin \left(x\right)}{2x}\right)
=\dfrac{-1-0}{2\pi}=-\dfrac{1}{2\pi} \quad (A)$$

 

[topsquark]

I am posting some AP calculus practice questions on MeWe so thot I would pass them thru here first
The solution is mine...
any typos or suggestions...

$\textbf{Find the Limit of}$
$\displaystyle\lim_{x\to \pi} \dfrac{\cos{x}+\sin{x}+1}{x^2-\pi^2}$
(A) $-\dfrac{1}{2\pi}$
(B) $\dfrac{1}{\pi}$
(C) $1$
(D) DNE
$\textbf{Solution}$
By observation we have $\frac{0}{0}$ so apply L'Hopital's Rule
apply LH'R then plug in $\pi$ and simplify
$$\displaystyle\lim _{x\to \:\pi }
\left(\frac{\cos \left(x\right)-\sin \left(x\right)}{2x}\right)
=\dfrac{-1-0}{2\pi}=-\dfrac{1}{2\pi} \quad (A)$$

Looks good to me.

-Dan