Question

What is the value of $\underset{x\to \infty }{\mathop{\lim }}\,{{x}^{2}}\sin \left( \dfrac{1}{x} \right)$ ?

Answer 1

Hint : We first change the variable of the limit with the conversion of $\dfrac{1}{x}=z$ . The limit value of the function also changes from $x\to \infty$ to $z\to 0$ . We change all the variables from $x$ to $z$ as $\underset{x\to \infty }{\mathop{\lim }}\,{{x}^{2}}\sin \left( \dfrac{1}{x} \right)=\underset{z\to 0}{\mathop{\lim }}\,\dfrac{\sin z}{{{z}^{2}}}$ . We use the theorems $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)g\left( x \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\times \underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$ and $\underset{z\to 0}{\mathop{\lim }}\,\dfrac{\sin z}{z}=1$ . We put the values to get $\underset{x\to \infty }{\mathop{\lim }}\,{{x}^{2}}\sin \left( \dfrac{1}{x} \right)=\infty$ .

Complete step by step solution:
First, we try to change the variable for the given limit value of $\underset{x\to \infty }{\mathop{\lim }}\,{{x}^{2}}\sin \left( \dfrac{1}{x} \right)$ .
We use the conversion of $\dfrac{1}{x}=z$ . The given limit form is $x\to \infty$ .
Due to the change of the variable the limit also changes to $z=\dfrac{1}{x}\to 0$ .
The value of the $x$ becomes $x=\dfrac{1}{z}$ .
The function changes to $\underset{x\to \infty }{\mathop{\lim }}\,{{x}^{2}}\sin \left( \dfrac{1}{x} \right)=\underset{z\to 0}{\mathop{\lim }}\,\dfrac{1}{{{z}^{2}}}\sin z=\underset{z\to 0}{\mathop{\lim }}\,\dfrac{\sin z}{{{z}^{2}}}$ .
Now we are going to apply the limit theorem of $\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)g\left( x \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)\times \underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$ .
Therefore, $\underset{z\to 0}{\mathop{\lim }}\,\dfrac{\sin z}{{{z}^{2}}}=\underset{z\to 0}{\mathop{\lim }}\,\left( \dfrac{\sin z}{z}\times \dfrac{1}{z} \right)=\underset{z\to 0}{\mathop{\lim }}\,\dfrac{\sin z}{z}\times \underset{z\to 0}{\mathop{\lim }}\,\dfrac{1}{z}$ .
Now we have the limit theorem $\underset{z\to 0}{\mathop{\lim }}\,\dfrac{\sin z}{z}=1$ .
We apply the theorem to get $\underset{z\to 0}{\mathop{\lim }}\,\dfrac{\sin z}{z}\times \underset{z\to 0}{\mathop{\lim }}\,\dfrac{1}{z}=\underset{z\to 0}{\mathop{\lim }}\,\dfrac{1}{z}$ .
Now we apply the limit value to get $\underset{z\to 0}{\mathop{\lim }}\,\dfrac{1}{z}=\infty$ .
Therefore, $\underset{x\to \infty }{\mathop{\lim }}\,{{x}^{2}}\sin \left( \dfrac{1}{x} \right)=\underset{z\to 0}{\mathop{\lim }}\,\dfrac{\sin z}{{{z}^{2}}}=\infty$ .
The limit value of $\underset{x\to \infty }{\mathop{\lim }}\,{{x}^{2}}\sin \left( \dfrac{1}{x} \right)$ is $\infty$ .
So, the correct answer is “ $\infty$”.

Note : The precise definition of a limit is something we use as a proof for the existence of a limit. When we’re evaluating a limit, we’re looking at the function as it approaches a specific point. we approach a particular value of x, the function itself gets closer and closer to a particular value.