Question

Find the limit. use l'hospital's rule if appropriate. if there is a more elementary method, consider using it. lim x→1 1 − x + ln(x) 1 + cos(5πx)

Answer 1

$\displaystyle\lim_{x\to1}\frac{1-x+\ln x}{1+\cos5\pi x}$

Evaluating the limand directly at $x=1$ gives an indeterminate form $\dfrac00$. Apply L'Hospital's rule once and we get

$\displaystyle\lim_{x\to1}\frac{-1+\frac1x}{-5\pi\sin5\pi x}$

Again, plugging in $x=1$ returns $\dfrac00$. Apply the rule once more:

$\displaystyle\lim_{x\to1}\frac{-\frac1{x^2}}{-25\pi^2\cos5\pi x}=\frac1{25\pi^2}\lim_{x\to1}\frac1{x^2\cos5\pi x}$

Now, in the denominator, when $x=1$ we get $x^2\cos5\pi x=-1$, so the limit is $-\dfrac1{25\pi^2}$.