[tex]\displaystyle\lim_{x\to1}\frac{1-x+\ln x}{1+\cos5\pi x}[/tex]
Evaluating the limand directly at [tex]x=1[/tex] gives an indeterminate form [tex]\dfrac00[/tex]. Apply L'Hospital's rule once and we get
[tex]\displaystyle\lim_{x\to1}\frac{-1+\frac1x}{-5\pi\sin5\pi x}[/tex]
Again, plugging in [tex]x=1[/tex] returns [tex]\dfrac00[/tex]. Apply the rule once more:
[tex]\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}[/tex]
Now, in the denominator, when [tex]x=1[/tex] we get [tex]x^2\cos5\pi x=-1[/tex], so the limit is [tex]-\dfrac1{25\pi^2}[/tex].
