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eimsori [14]
3 years ago
5

Lim x → 1 ln(x)/ sin(7πx)

Mathematics
1 answer:
mote1985 [20]3 years ago
7 0
\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}
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