Question

Lim
x → 1
ln(x)/
sin(7πx)

Answer 1

`\displaystyle\lim_(x\to1)(\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)(\lnx )/(\sin7\pi x)=\lim_(x\to1)(\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}`