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

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By l'Hopital's Rule, we can find

Let us look at some details.

by rewriting a little bit,

by l'Hopital's Rule,

by cancelling out

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You could also easily find an answer by looking at the known limit of

In your problem, changing working variable makes it simpler: in fact, if we assume

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Instead of l'Hopital's Rule, one can use the fundamental trigonometric limit:

The limit you are interested in can be written:

Now, as

With

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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When you substitute in infinity,

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

**Applying L'Hopital**

**Simplify the previous step**

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**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(A*B) = 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

Let

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

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