Question

The value of limit Limx→0 {sin (a + x) – sin (a – x)}/x is​

Answer 1

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$\large \boxed{ \red{\tt :  ⟼\lim_{x\to 0}\dfrac{sinx}{x} \: = 1}}$

$\large \boxed{ \green{\tt :  ⟼sinx \: - siny \: = 2cos(\dfrac{x + y}{2} )sin(\dfrac{x - y}{2} )}}$

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$\large\underline\purple{\bold{Solution :- }}$

$\tt :  ⟼\lim_{x\to 0} \: \dfrac{sin(a + x) - sin(a - x)}{x}$

☆ On substituting directly x = 0, we get indeterminant form

$\tt :  ⟼ \: \lim_{x\to 0}\dfrac{2cos \bigg(\dfrac{a + x + a - x}{2} \bigg) \: sin\bigg(\dfrac{a + x - a + x}{2} \bigg)}{x}$

$\tt :  ⟼ \: 2 \: \lim_{x\to 0}\dfrac{cos\bigg(\dfrac{2a}{2} \bigg)sin\bigg(\dfrac{2x}{2} \bigg)}{x}$

$\tt :  ⟼ \: 2\lim_{x\to 0} \: \dfrac{cosa \: sinx}{x}$

$\tt :  ⟼ \: 2 \: cosa \: \lim_{x\to 0} \: \dfrac{sinx}{x}$

$\tt :  ⟼ \: 2 \: cosa \: \times 1$

$\tt :  ⟼ \: 2 \: cosa$

$\boxed{ \red{\bf \: Hence \: \tt :  ⟼\lim_{x\to 0} \: \dfrac{sin(a + x) - sin(a - x)}{x} = 2cosa}}$

Answer 2

Answer:

(c) 2 cosa

Step-by-step explanation:

Given, Limx→0 {sin (a + x) – sin (a – x)}/x

= Limx→0 {2 × cos a × sin x}/x

= 2 × cos a × Limx→0 sin x/x

= 2 cos a