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Find the limit. use l'hospital's rule if appropriate. lim x→infinity x(tan 7/x

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1 vote

\displaystyle\lim_(x\to\infty)x\tan\frac7x=7\lim_(x\to\infty)(\sin\frac7x)/(\frac7x\cos\frac7x)

Replace
y=\frac7x. Then as
x\to\infty, you have
y\to0^+, so the limit is equivalent to


\displaystyle7\lim_(y\to0^+)(\sin y)/(y\cos y)=7\left(\lim_(y\to0^+)\frac{\sin y}y\right)\left(\lim_(y\to0^+)\frac1{\cos y}\right)

The first limit is well-known and has a value of 1. Meanwhile
\cos y is continuous at
y=0 so the second limit evaluates to
\frac1{\cos0}=1.

This means the limit must be 7.

No L'Hopital's rule needed!
User Cheyanne
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