Question

evaluate the following limit​

Answer 1

EXPLANATION.

$\implies \displaystyle \lim_{x \to 0} \bigg(\dfrac{1}{x^{2} } - \dfrac{1}{sin^{2}x } \bigg)$

As we know that,

We can write equation as,

$\implies \displaystyle \lim_{x \to 0} \bigg(\dfrac{sin^{2} x - x^{2} }{x^{2} (sin^{2} x)} \bigg)$

Put the value of x = 0 in the equation and check their indeterminant form, we get.

$\implies \displaystyle \lim_{x \to 0} \bigg(\dfrac{sin^{2}(0) - (0)^{2} }{(0)^{2} (sin^{2} (0))} \bigg) \ = \dfrac{0}{0}$

As we can see that it is a 0/0 form of indeterminant.

As we know that,

Standard expansion.

$\implies \displaystyle \lim_{x \to 0} \bigg(\dfrac{sin(x)}{x} \bigg) = \displaystyle \lim_{x \to 0} \bigg(\dfrac{x}{sin(x)} \bigg) = 1$

Using this expansion in the equation, we get.

Multiply and divide numerator and denominator by x², we get.

$\implies \displaystyle \lim_{x \to 0} \bigg(\dfrac{sin^{2} x - x^{2} }{x^{2} sin^{2}x } \bigg) \times \dfrac{x^{2} }{x^{2} }$

$\implies \displaystyle \lim_{x \to 0} \bigg(\dfrac{sin^{2} x - x^{2} }{x^{4} } \bigg) \times \displaystyle \lim_{x \to 0} \bigg(\dfrac{x^{2} }{sin^{2} x} \bigg)$

$\implies \displaystyle \lim_{x \to 0} \bigg(\dfrac{sin^{2} x - x^{2} }{x^{4} } \bigg) \times \displaystyle \lim_{x \to 0} \bigg(\dfrac{x^{} }{sin{} x} \bigg)^{2}$

$\implies \displaystyle \lim_{x \to 0} \bigg(\dfrac{sin^{2} x - x^{2} }{x^{4} } \bigg)$

Now, again put the value of x = 0 in the equation, we get.

$\implies \displaystyle \lim_{x \to 0} \bigg(\dfrac{sin^{2}(0) - (0)^{2} }{(0)^{4} } \bigg) \ = \dfrac{0}{0}$

Again, we see that it is in the form of 0/0 indeterminant form, we get.

Now, apply L-HOSPITAL'S rule in this equation, we get.

Apply L-HOSPITAL'S till there denominator of power comes to = 1.

$\implies \displaystyle \lim_{x \to 0} \bigg(\dfrac{2sin(x).cos(x) - 2x}{4x^{3} } \bigg)$

$\implies \displaystyle \lim_{x \to 0} \bigg(\dfrac{sin(2x) - 2x}{4x^{3} } \bigg)$

Again, Apply L-HOSPITAL'S rule, we get.

$\implies \displaystyle \lim_{x \to 0} \bigg(\dfrac{2cos(2x) - 2}{12x^{2} }\bigg)$

$\implies \displaystyle \lim_{x \to 0} \bigg( \dfrac{2(cos2x - 1)}{12x^{2} } \bigg)$

$\implies \displaystyle \lim_{x \to 0} \bigg(\dfrac{cos2x - 1}{6x^{2} } \bigg)$

As we know that,

Formula of :

⇒ Cos2x = 1 - 2sin²x.

Using this formula in the equation, we get.

$\implies \displaystyle \lim_{x \to 0} \bigg(\dfrac{1 - 2sin^{2}x - 1 }{6x^{2} } \bigg)$

$\implies \displaystyle \lim_{x \to 0} \bigg( \dfrac{- 2sin^{2}x }{6x^{2} } \bigg)$

$\implies \displaystyle \lim_{x \to 0} \dfrac{-2}{6} \bigg(\dfrac{sin^{2}x }{x^{2} } \bigg)$

$\implies \displaystyle \lim_{x \to 0} \dfrac{-1}{3} \bigg(\dfrac{sinx}{x} \bigg)^{2} = \dfrac{-1}{3}$

$\implies \displaystyle \lim_{x \to 0} \bigg(\dfrac{1}{x^{2} } - \dfrac{1}{sin^{2}x } \bigg) = \dfrac{-1}{3}$