\(\displaystyle\lim_{x\to\infty}x\tan\frac7x=7\lim_{x\to\infty}\frac{\sin\frac7x}{\frac7x\cos\frac7x}\)
Replace \(y=\dfrac7x\). Then as \(x\to\infty\), you have \(y\to0^+\), so the limit is equivalent to
\(\displaystyle7\lim_{y\to0^+}\frac{\sin y}{y\cos y}=7\left(\lim_{y\to0^+}\frac{\sin y}y\right)\left(\lim_{y\to0^+}\frac1{\cos y}\right)\)
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 \(\dfrac1{\cos0}=1\).
This means the limit must be 7.
No L'Hopital's rule needed!
