Question

please solve the question
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Answer 1

$\displaystyle\\\lim_{x\to a}\dfrac{x\sin a -a \sin x}{x-a}=\\\\\lim_{x\to a}\dfrac{(x\sin a -a \sin x)'}{(x-a)'}=\\\\\lim_{x\to a}\dfrac{\sin a -a\cos x}{1}=\\\\\lim_{x\to a}(\sin a -a\cos x)=\\\\\sin a-a \cos a$

Answer 2

Answer:

$\large\bold\red{ \sin(a) - a \cos(a)}$

Step-by-step explanation:

Given,

$\lim_{x\to \: a} \frac{x \sin(a) - a \sin(x) }{x - a} </p><p>$

If we put x = a, the values comes 0/0.

now,

since it is in the form of 0/0.

therefore,

we can apply L - Hospitals Rule here.

• In L - Hospitals Rule, we differentiate both Numerator and Denominator respectively.

Thus,

we get,

$\lim_{x\to \: a} \frac{ \sin(a) \frac{d}{dx} x - a \frac{d}{dx} \sin(x) }{ \frac{d}{dx}x - \frac{d}{dx} a} \\ \\ = \lim_{x\to \: a} \frac{ \sin(a) - a \cos(x) }{1 - 0} \: \: \\ \\ = \bold{ \sin(a) - a \cos(a)} </p><p></p><p>$