Proving $\lim_{x\to 0}\dfrac{\sin x}{x} = 1$ using Laplace

Question

I was trying to prove below equation using different way.

$$\lim_{x \to 0} \frac{\sin x}x = 1$$

One of the ways I though was using Laplace (given below) but unable to prove it? Kindly help me to prove this equation using Laplace without taking Laplace inverse.

1.Using initial value theorem: $\lim_{t \to 0}y(t) = \lim_{s \to \infty}sY(s)$.

Using Laplace Transform $$\lim_{x \to 0} \frac{\sin x}x = \lim_{s \to \infty} s\; \mathcal{L}\left(\frac{\sin x}x\right) = \lim_{s \to \infty} s \int_{s}^{\infty} \frac{1}{s^2+1}\,ds = \lim_{s \to \infty} s \left(\frac{\pi}2 - \arctan s\right) $$

I am stuck at this step, I don't know how to proceed forward this.

Thanks.

Answer 1

Write $$\lim_{s\to\infty}s\left(\frac{\pi}{2}-\arctan s\right) = \lim_{s\to\infty} \frac{\pi/2 - \arctan s}{1/s}$$ and then use De L'Hospital to calculate the latter limit which is of the form $0/0$.