Question

Evaluate the following limit:-

$\sf\displaystyle \lim_{x \to 0} \dfrac{\sin\bigg(\pi\cos^{2}x\bigg) }{x^{2} }$

No go ogled contents please!

Answer 1

Answer:

$\huge{\pi}$

Step-by-step explanation:

Given :

$\mathrm{\lim \limits_{x \rightarrow 0} \dfrac{sin\bigg(\pi cos^2x\bigg)}{x^2}}$

To find :

The value of the given limit.

Solution :

$\longmapsto \mathrm{\lim \limits_{x \rightarrow 0} \dfrac{sin\bigg(\pi cos^2x\bigg)}{x^2}}$

We know that,

$\boxed{\mathrm{cos^2x = 1 - sin^2x}}$

$\implies \mathrm{\lim \limits_{x \rightarrow 0} \dfrac{sin\bigg(\pi (1-sin^2x)\bigg)}{x^2}}$

$\implies \mathrm{\lim \limits_{x \rightarrow 0} \dfrac{sin\bigg(\pi - \pi sin^2x\bigg)}{x^2}}$

We know that,

$\boxed{\mathrm{sin(\pi - x) = sinx}}$

$\implies \mathrm{\lim \limits_{x \rightarrow 0} \dfrac{sin\bigg(\pi sin^2x\bigg)}{x^2}}$

Now multiplying and dividing numerator and denominator with πsin²x

$\implies \mathrm{\lim \limits_{x \rightarrow 0} \dfrac{sin\bigg(\pi sin^2x\bigg)}{x^2} \times \dfrac{\pi sin^2x}{\pi sin^2x}}$

Rearranging, we get,

$\implies \mathrm{\lim \limits_{x \rightarrow 0} \dfrac{sin\bigg(\pi sin^2x\bigg)}{\pi sin^2x} \times \dfrac{\pi sin^2x}{x^2}}$

We know that,

$\boxed{\mathrm{\lim \limits_{h \rightarrow o} \dfrac{sinh}{h} = 1}}$

Here, we have 2 limits with πsin²x and x² in place of h

So,

$\implies \mathrm{\Bigg(\lim \limits_{x \rightarrow 0} \dfrac{sin\bigg(\pi sin^2x\bigg)}{\pi sin^2x}\Bigg) \times \pi \times \Bigg(\lim \limits_{x \rightarrow 0} \dfrac{sin^2x}{x^2}\Bigg)}$

$\implies{\mathrm{1 \times \pi \times 1}}$

$\implies \underline{\underline{\mathrm{\pi}}}$

$\therefore \underline{\boxed{\mathbf{\lim \limits_{x \rightarrow 0} \dfrac{sin\bigg(\pi cos^2x\bigg)}{x^2} = \pi}}}$

Hope it helps!!

Answer 2

The movement of the body or object or a particle, that is following a circular path is called a circular motion. Now, the motion of a body or object or a particle moving with constant speed along a circular path is called Uniform Circular Motion.