$\displaystyle\lim_{n\to\infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}} = 0$, where is my mistake?

Question

$\displaystyle\lim_{n\to\infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}} = 1$, but when I evaluate it as follows I get $0$ as the result:

\begin{align} \lim_{n\to\infty} \frac{\sin \frac{1}{n}}{\frac{1}{n}} & = \lim_{n\to\infty} \left(n \sin \frac{1}{n} \right) \\ & = \lim_{n\to\infty} \left( \frac{n \left (\sin \frac{1}{n} \right) ^2}{\sin \frac{1}{n}} \right) \\ & = \lim_{n\to\infty} \left( \left( \sin \frac{1}{n} \right) ^2 \frac{n}{\sin \frac{1}{n}} \right) \\ & = \lim_{n\to\infty} \left( \sin \frac{1}{n} \right) ^2 \lim_{n\to\infty} \frac{n}{\sin \frac{1}{n}} \\ & \underset{\text{L'Hôpital}}{=} \left( \sin \left( \lim_{n\to\infty} \frac{1}{n} \right) \right) ^2 \lim_{n\to\infty} \frac{1}{\cos \frac{1}{n}} \\ & = \left( \sin 0 \right) ^2 \frac{1}{\cos \left( \displaystyle\lim_{n\to\infty} \frac{1}{n} \right) } \\ & = 0^2 \frac{1}{\cos 0} \\ & = 0 \cdot \frac{1}{1} \\ & = 0 \cdot 1 \\ & = 0 \end{align}

Where is my mistake?

Thank you.

Answer 1

You cannot use L'Hopital's rule to evaluate the limit $$\lim_{n\to\infty} \frac{n}{\sin \frac{1}{n}},$$ as it is not of the form $\frac{\to 0}{\to 0}$ or $\frac{\to \infty}{\to \infty}.$ In fact, the fraction $\frac{n}{\sin \frac{1}{n}}$ approaches $\infty$ as $n\to\infty$. Since this limit does not exist (i.e., it is not finite) - the splitting of the limit of the product of two functions into a product of limits is not justified.

Answer 2

Where is my mistake?

You cannot say $$ \lim_{n\to\infty} \left( \left( \sin \frac{1}{n} \right) ^2 \frac{n}{\sin \frac{1}{n}} \right) = \lim_{n\to\infty} \left( \sin \frac{1}{n} \right) ^2 \lim_{n\to\infty} \frac{n}{\sin \frac{1}{n}},$$ because the limit of the products equals the product of the limits when the limits are defined,

and $$\lim\limits_{n\to\infty} \frac{n}{\sin \frac{1}{n}}=\infty.$$

Answer 3

As noticed we can't do this step

$$\ldots= \lim_{n\to\infty} \left( \left( \sin \frac{1}{n} \right) ^2 \frac{n}{\sin \frac{1}{n}} \right) = \lim_{n\to\infty} \left( \sin \frac{1}{n} \right) ^2 \lim_{n\to\infty} \frac{n}{\sin \frac{1}{n}}=\ldots$$

because you have factorized in an indeterminate form $0 \cdot \infty$ which is not allowed, as for example in the case

$$1=\lim_{n\to \infty}\frac n n \neq \lim_{n\to \infty}n\cdot \lim_{n\to \infty}\frac 1 n $$