# find $\lim_{x\to 1} \frac{\ln x - x + 1}{e^{\pi(x-1)} + \sin (\pi x) -1}$

(a) $\frac{-1}{\pi}$ -------- (b)$\frac{-1}{\pi-1}$

(c) $\frac{-1}{\pi^2}$ -------- (d) undefined

You obviously can't solve the limit straightforwardly, because it gives you an indeterminate form (0/0), right? So I used L'Hôpital's rule, taking the derivative of the numerator and the denominator before plugging in x=1.

Derivative of the numerator: $\frac1x -1$

Derivative of the denominator: $\pi(e^{\pi(x-1)} + \cos(\pi x))$

However, the numerator remains unavoidably equal to zero. Am I doing something wrong, glancing over something really obvious, or are the answer choices wrong? Any help would be greatly appreciated.

• well e^pi + 1 = 0, which may get you somewhere – Alex Robinson Dec 5 '16 at 19:19

No. One may apply L'Hospital's rule once more obtaining, as $x \to 1$, $$\frac{\left(\frac1x-1\right)'}{\left(\pi(e^{\pi(x-1)} + \cos(\pi x))\right)'}=\frac{-\frac1{x^2}}{\pi^2(e^{\pi(x-1)} -\sin(\pi x))}\to-\frac1{\pi^2}.$$
• You mean $\frac{0}{\pi (e^0-1)}$ which is $\frac{0}{0}$. Isn't it? – Olivier Oloa Dec 5 '16 at 19:48