What is the value of \(\rm \displaystyle \lim_{x \to 0} {x^2}{e^{\sin \left( {\tfrac{1}{x}} \right)}}\)?

Concept:

The Squeeze Theorem (The Sandwich Theorem): is used on a function where it will be almost impossible to differentiate.

• The squeeze theorem states that if we define functions such that h(x) ≤ f(x) ≤ g(x) and if \(\rm \displaystyle \lim_{x \to a}h(x) = \lim_{x \to a}g(x) = L\), then \(\rm \displaystyle \lim_{x \to a}f(x) = L\).

Calculation:

We know that -1 ≤ sin θ ≤ 1.

⇒ -1 ≤ \(\rm \sin \left(\dfrac1x\right)\) ≤ 1

Since, e^x is a strictly increasing function for all real values of x, we can say that:

⇒ e^-1 ≤ \(\rm e^{\sin \left(\tfrac1x\right)}\) ≤ e¹

Also, since x² ≥ 0, we can say that:

⇒ x²e-1 ≤ \(\rm x^2e^{\sin \left(\tfrac1x\right)}\) ≤ x²e1

⇒ \(\rm \dfrac{x^2}{e}\leq x^2e^{\sin \left(\tfrac1x\right)}\leq x^2e\)

So, we can consider h(x) = \(\rm \dfrac{x^2}{e}\), f(x) = \(\rm \displaystyle \lim_{x \to 0} {x^2}{e^{\sin \left( {\tfrac{1}{x}} \right)}}\) and g(x) = x²e.

Now, \(\rm \displaystyle \lim_{x \to 0}h(x) = \lim_{x \to 0}\dfrac{x^2}{e}=0\).

And \(\rm \displaystyle \lim_{x \to 0}g(x) = \lim_{x \to 0}x^2e=0\).

Since, \(\rm \displaystyle \lim_{x \to 0}h(x) =\lim_{x \to 0}g(x) =0\), we must have \(\rm \displaystyle \lim_{x \to 0}f(x) =0\).

Hence, \(\rm \displaystyle \lim_{x \to 0} {x^2}{e^{\sin \left( {\tfrac{1}{x}} \right)}}=0\).