# What is the limit of the function sin x/squareroot(x^2+1), x->+infinite?

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### 2 Answers

We have to find the value of lim x--> inf+ [sin x / sqrt (x^2 + 1)]

sin x is a periodic function the value of which oscillates between -1 and 1 as x tends to infinity. It is not possible to determine the value of sin x for x--> inf., but we know the value is between -1 and +1.

The value of sqrt (1 + x^2) tends to infinity as x tends to inf. So the value of 1/ sqrt ( x^2 + 1) tends to zero.

This allows us to say that the value of lim x--> inf+ [sin x / sqrt (x^2 + 1)] = 0

**The required value of lim x--> inf+ [sin x / sqrt (x^2 + 1)] is 0.**

### User Comments

We cannot calculate the limit of sine function if x-> +infinite.

We'll calculate the function based on the property of product of 2 functions, if the limit of one function is 0 and the other function is bordered.

By definition, the sine function is limited by the values -1 and 1.

|sin x|=<1

Now, we'll calculate the limit of the fraction 1/sqrt(x^2+1):

lim 1/sqrt(x^2+1) = 1/infinite = 0

We'll write the given function as the product of 2 functions:

**lim [1/sqrt(x^2+1)]*(sin x) = 0, if x->infinite**