Respuesta :

LammettHash
[tex]\displaystyle\lim_{x\to\infty}x\tan\frac7x=7\lim_{x\to\infty}\frac{\sin\frac7x}{\frac7x\cos\frac7x}[/tex]

Replace [tex]y=\dfrac7x[/tex]. Then as [tex]x\to\infty[/tex], you have [tex]y\to0^+[/tex], so the limit is equivalent to

[tex]\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)[/tex]

The first limit is well-known and has a value of 1. Meanwhile [tex]\cos y[/tex] is continuous at [tex]y=0[/tex] so the second limit evaluates to [tex]\dfrac1{\cos0}=1[/tex].

This means the limit must be 7.

No L'Hopital's rule needed!

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