Question

lim
x → 1
ln(x)/
sin(7πx)

Answer 1

$\displaystyle\lim_{x\to1}\frac{\ln x}{\sin7\pi x}$

Note that both the numerator and denominator approach 0 as $x\to1$, so we can try using L'Hopital's rule.

$\displaystyle\lim_{x\to1}\frac{\lnx }{\sin7\pi x}=\lim_{x\to1}\frac{\frac1x}{7\pi\cos7\pi x}=\lim_{x\to1}\frac1{7\pi x\cos7\pi x}$

The denominator is nonzero at $x=1$, so the limit is equivalent to

$\displaystyle\frac1{\lim\limits_{x\to1}7\pi x\cos7\pi x}=\frac1{7\pi\cos7\pi}=-\frac1{7\pi}$