Question

Somebody Please solve this question without using L Hospital rule.Evaluate if:​

Answer 1

Answer:

1/2

Step-by-step explanation:

Given that,

$\lim_{n \to 1} (\dfrac{1+\cos\pi x}{\tan^2\pi x})$

Using L Hospital's rule to find it :

$\lim_{n \to a} \dfrac{f(x)}{g(x)}= \lim_{n \to a} \dfrac{f'(x)}{g'(x)}$

We have,

a = 1, $f(x)=1+\cos\pi x, \ \ g(x)=\tan^2\pi x$

$f'(x)=\dfrac{d}{dx}(1+\cos\pi x)\\\\=-\pi \sin \pi x\ .....(1)$

$g'(x)=\dfrac{d}{dx}(\tan^2\pi x)\\\\=2\tan\pi x\times \sec^2\pi x\times \pi\ .....(2)$

From equation (1) and (2) :

$\lim_{n \to 1} \dfrac{f(x)}{g(x)}= \lim_{n \to 1} \dfrac{-\pi \sin\pi x}{2\tan \pi x\times \sec^2\pi x \times \pi}\\\\\lim_{n \to 1} \dfrac{-\pi \sin\pi x}{2\tan \pi x\times \sec^2\pi x \times \pi}\\\\=\lim_{n \to 1} \dfrac{-1}{2}\times\cos^3\pi x\\\\=\dfrac{-1}{2}\times \cos^3\pi \\\\=\dfrac{1}{2}$

So, the value of the given function is 1/2.