Question

Find the value of $\underset{x\to 0}{lim}{x}^{2}sin\left(\frac{1}{x}\right)$ using sandwich theorem.

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

We have,$\underset{x\to 0}{lim}{x}^{2}sin\left(\frac{1}{x}\right)$

When x get closer to zero, the function f(x)=$sin\left(\frac{1}{x}\right)$ fails to have limit.so, we are not able to use basic prpoperties of substituting directly x=0 to the limit.

But we know that this function f(x) = $sin\left(\frac{1}{x}\right)$ is bounded by -1 and above 1.

i.e., -1$\le$ $sin\left(\frac{1}{x}\right)$ $\le 1$

For any real number x

$x^2\ge 0$

We can mulitiply $x^2$ on both side of inequality.

$-x^2\le x^2$ $sin\left(\frac{1}{x}\right)$ $\le x^2$

Taking $\underset{x\to 0}{lim}\underset{h\left(x\right)}{\underset{\downarrow }{(-x^2)}}\le \underset{x\to 0}{lim}\underset{f\left(x\right)}{\underset{\downarrow }{x^{2}sin\left(\frac{1}{x}\right)}}\le \underset{x\to 0}{lim}\underset{g\left(x\right)}{\underset{\downarrow }{\left(x^2\right)}}$

According to sandwich theorem

If $h\left(x\right)\le f\left(x\right)\le g\left(x\right)$

$\underset{x\to a}{lim}h\left(x\right)=Land\underset{x\to a}{lim}g\left(x\right)=L$