1

How do I find the limit:

$$\lim_{x\to\infty}x\sin(\tan\frac1x)$$

Ivan Neretin
  • 13,429
Dole
  • 2,754
  • 1
    Change of variable and l'Hopital? Please add some context to this bare problem statement question to avoid the impression you are simply paasing through an assigned exercise to your Readers. – hardmath Feb 22 '17 at 12:31

2 Answers2

5

Set $x=\frac{1}{t}$. Note that $$\lim_{x \to \infty} x \sin \left(\tan \frac{1}{x}\right)=\lim_{t \to 0^{+}} \frac{\sin (\tan t)}{t}$$ Now, note that $$\lim_{t \to 0^{+}} \frac{\sin (\tan t)}{t}=\lim_{t \to 0^{+}} \frac{\sin (\tan t)}{\tan t} \times \frac{\tan t}{t}=\lim_{t \to 0^{+}} \frac{\sin (\tan t)}{\tan t} \times \lim_{t \to 0^{+}}\frac{\tan t}{t}$$ Setting $\tan t =u$, the limit becomes $$\lim_{u \to 0^{+}} \frac{\sin u}{u} \times \lim_{t \to 0^{+}} \frac{\sin t}{t} \times \lim_{t \to 0^{+}} \frac{1}{\cos t}=1$$ As proven here.

Community
  • 1
S.C.B.
  • 23,106
0

Rewrite the limit as

\begin{align} L:=\lim\limits_{x\rightarrow\infty}x\sin\left(\tan\frac{1}{x}\right)=\lim\limits_{x\rightarrow\infty}\frac{\sin\left(\tan\frac{1}{x}\right)}{\frac{1}{x}} \end{align}

Since both the numerator and the denominator tend to $0$ in the limit, this is an indeterminate form and we can apply L'Hospital's Rule.

\begin{align} L&=\lim\limits_{x\rightarrow\infty}\frac{\sin\left(\tan\frac{1}{x}\right)}{\frac{1}{x}}\\ &=\lim\limits_{x\rightarrow\infty}\frac{-\frac{1}{x^2}\sec^2\left(\frac{1}{x}\right)\cos\left(\tan\frac{1}{x}\right)}{-\frac{1}{x^2}}\\ &=\lim\limits_{x\rightarrow\infty}\sec^2\left(\frac{1}{x}\right)\cos\left(\tan\frac{1}{x}\right)\\ \end{align}

By continuity of all the functions involved, we can pull the limits into the functions. Since $\frac{1}{x}\rightarrow 0$ as $x\rightarrow\infty$, we have

$$\lim\limits_{x\rightarrow\infty}x\sin\left(\tan\frac{1}{x}\right) = 1$$

Tom
  • 3,319