Question

Please solve this question please
Chapter - Limits and derivatives
Please don't spam. I am doing test analysis, very important so don't spam please​ Don't use L Hospital's rule please

Answer 1

EXPLANATION.

$\sf \implies \displaystyle \lim_{x \to a} \bigg( \dfrac{cos(x) - cos(\alpha )}{cot(x) - cot(\alpha )} \bigg)$

As we know that,

First we put the value of x = a in the equation and check their indeterminant form, we get.

$\sf \implies \displaystyle \lim_{x \to a} \bigg( \dfrac{cos(\alpha ) - cos(\alpha )}{cot(\alpha ) - cot(\alpha )} \bigg) = \dfrac{0}{0} \ form$

It is 0/0 form indeterminant, we get.

Differentiate numerator and denominator of the equation, we get.

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

$\sf \implies \displaystyle \lim_{x \to a} sin^{3} (x)$

Put the value of x = a in the equation, we get.

⇒ sin³a.

$\sf \implies \displaystyle \lim_{x \to a} \bigg( \dfrac{cos(x) - cos(\alpha )}{cot(x) - cot(\alpha )} \bigg) = sin^{3} a$

Option [A] is correct answer.

Answer 2

I'll use the limits at the end, because I'm not so comfortable with LaTeX, sorry.

So, since you have restricted the use of L'H rule,let me take the long way and solve your question.

→ cos(x) - cos(a)/[(cos(x)/sin (x) - cos(a)/sin (a)] (°•° representing cot as cos/sin)

→ cos(x) - cos(a)/[(cos(x)sin(a) - cos(a)sin (x))/sin (x) sin(a)]

→ (cos (x) - cos(a))•(sin (x) sin(a))/cos(x) sin(a) - cos(a) sin (x)

→ (cos (x) - cos (a)) • (sin (x) sin (a))/sin (a-x)

[°•° sin (A) cos(B) - cos(A) sin (B) = sin (A-B)]

→ (cos(x) - cos(a))(sin (x) sin (a))/ sin (a/2 - x/2)

→ (2 sin [(x+a)/2] • sin [(a-x)/2]) (sin (a) sin (x))/2 sin [(a-x)/2] • cos [(x-a)/2]

→ 2 sin [(x+a)/2] (sin (a) sin (x)/2 cos [(x-a)]/2 (°•° Cancelling the terms from Nr and Dr)

Substitute the limit x → a

→ sin [(a+a)/2] (sin (a) sin (a)/ cos [(a-a)/2]

→ sin (2a)/2 (sin²a)/cos (0)

→ sin (a) (sin²(a)/1

→ sin (a) • sin²(a)

→ sin³a

Option (A) is correct.