# Find the limit as x approaches infinity of xsin(1/x)?

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126
Nov 8, 2016

By l'Hopital's Rule, we can find
${\lim}_{x \to \infty} x \sin \left(\frac{1}{x}\right) = 1$.

Let us look at some details.
${\lim}_{x \to \infty} x \sin \left(\frac{1}{x}\right)$

by rewriting a little bit,
$= {\lim}_{x \to \infty} \frac{\sin \left(\frac{1}{x}\right)}{\frac{1}{x}}$

by l'Hopital's Rule,
$= {\lim}_{x \to \infty} \frac{\cos \left(\frac{1}{x}\right) \cdot \left(- \frac{1}{x} ^ 2\right)}{- \frac{1}{x} ^ 2}$

by cancelling out $- \frac{1}{x} ^ 2$,
$= {\lim}_{x \to \infty} \cos \left(\frac{1}{x}\right) = \cos \left(0\right) = 1$

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Tomasz Share
Jun 12, 2015

You could also easily find an answer by looking at the known limit of $\sin x$:

${\lim}_{t \rightarrow 0} \frac{\sin t}{t} = 1$

In your problem, changing working variable makes it simpler: in fact, if we assume $t = \frac{1}{x}$, $t \rightarrow 0$ for $x \rightarrow \infty$ and therefore with this simple substitution you have ${\lim}_{x \rightarrow \infty} x \sin \left(\frac{1}{x}\right) = {\lim}_{x \rightarrow \infty} \sin \frac{\frac{1}{x}}{\frac{1}{x}} = {\lim}_{t \rightarrow 0} \sin \frac{t}{t} = 1$

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Jim H Share
Apr 8, 2015

Instead of l'Hopital's Rule, one can use the fundamental trigonometric limit: ${\lim}_{h \rightarrow 0} \sin \frac{h}{h} = 1$.

The limit you are interested in can be written: ${\lim}_{x \rightarrow \infty} \sin \frac{\frac{1}{x}}{\frac{1}{x}}$.

Now, as $x \rightarrow \infty$, we know that $\frac{1}{x} \rightarrow 0$ and we can think of the limit as

${\lim}_{\frac{1}{x} \rightarrow 0} \sin \frac{\frac{1}{x}}{\frac{1}{x}}$.

With $h = \frac{1}{x}$, this becomes${\lim}_{h \rightarrow 0} \sin \frac{h}{h}$ which is $1$.

Although it is NOT needed, here's the graph of the function:

graph{y = x sin(1/x) [-5.55, 5.55, -2.775, 2.774]}

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14
Jun 6, 2015

When you substitute in infinity, $\infty$, you end up with the indeterminate form of $\infty \cdot 0$.

${\lim}_{x \to \infty} x \sin \left(\frac{1}{x}\right) = \infty \cdot \sin \left(\frac{1}{\infty}\right) = \infty \cdot \sin \left(0\right) = \infty \cdot 0$

We still have options though. We now can fall back on L'Hopital's Rule which basically says to take the derivative of the numerator and denominator independently. Do not use the quotient rule.

We need to rewrite this function so that is produces an indeterminate in the form $\frac{\infty}{\infty}$ or $\frac{0}{0}$.

${\lim}_{x \to \infty} \frac{\sin \left(\frac{1}{x}\right)}{x} ^ \left(- 1\right) = \sin \frac{\frac{1}{x}}{\frac{1}{x}} = \sin \frac{\frac{1}{\infty}}{\frac{1}{\infty}} = \sin \frac{0}{0} = \frac{0}{0}$

Applying L'Hopital

${\lim}_{x \to \infty} \frac{\left(\sin \left(\frac{1}{x}\right)\right) '}{\left({x}^{- 1}\right) '}$

$= {\lim}_{x \to \infty} \frac{\left(- 1\right) \cdot {x}^{- 2} \cdot \cos \left(\frac{1}{x}\right)}{\left(- 1\right) \cdot {x}^{- 2}}$

Simplify the previous step

$= {\lim}_{x \to \infty} \cos \left(\frac{1}{x}\right) = \cos \left(\frac{1}{\infty}\right) = \cos \left(0\right) = 1$

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5
Jul 5, 2015

Original:
Answer is that there is no limiting value/ limit does not exist

It is because the value of sin1/x does not have a limiting value and oscillates between 0 and 1 while the value of x does not have a limit but increases as we increase the number. Thus according to the law lim(AB) = lim(A)lim(B), our answer is justified.

To help ur understanding:From the basic concept of limit, we increase the value of x from a large number say 10^3 and increase it to 10^4 and so on(towards infinity), we see no limiting value. Thus we get the answer.

Revised:
As others have suggested, L'Hopital's Rule certainly works. This is of the form $\infty \cdot 0$, which is indeterminate. We can clearly rewrite it like this:

${\lim}_{x \to \infty} \sin \frac{\frac{1}{x}}{\frac{1}{x}}$

Let $\frac{1}{x} = u$:

$= {\lim}_{\frac{1}{u} \to \infty} \sin \frac{u}{u}$

$= {\lim}_{u \to 0} \sin \frac{u}{u}$

Thus with L'Hopital's Rule and the Chain Rule for substitution:

$= {\lim}_{u \to 0} \frac{\cos u \cdot \cancel{\left(\frac{\mathrm{du}}{\mathrm{dx}}\right)}}{1 \cancel{\left(\frac{\mathrm{du}}{\mathrm{dx}}\right)}}$

$= {\lim}_{u \to 0} \cos u = \cos 0 = 1$

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