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Question: Find the value of the limit \(\displaystyle \lim_{x \to a}\dfrac{x\sin a-a\sin x}{x-a}\)....

Find the value of the limit lim⁡x→axsin⁡a−asin⁡xx−a\displaystyle \lim_{x \to a}\dfrac{x\sin a-a\sin x}{x-a}.

Explanation

Solution

Add and subtract asina in the numerator. Hence prove that the given limit is equal to lim⁡x→a(x−a)sin⁡ax−a+a(sin⁡a−sin⁡x)x−a\displaystyle \lim_{x \to a}\dfrac{\left( x-a \right)\sin a}{x-a}+\dfrac{a\left( \sin a-\sin x \right)}{x-a}.
Use the fact that lim⁡x→af(x)−f(a)x−a=f′(a)\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)-f\left( a \right)}{x-a}=f'\left( a \right). Hence evaluate the limit of the given function.
Alternatively, us L.H rule to find the given limit

** Complete step-by-step answer :**
We have
l=lim⁡x→axsin⁡a−asin⁡xx−al=\displaystyle \lim_{x \to a}\dfrac{x\sin a-a\sin x}{x-a}
Adding and subtracting asina in the numerator of RHS, we get
l=lim⁡x→axsin⁡a−asin⁡x−asin⁡a+asin⁡ax−al=\displaystyle \lim_{x \to a}\dfrac{x\sin a-a\sin x-a\sin a+a\sin a}{x-a}
Re-writing the terms in the numerator in different order, we get
l=lim⁡x→axsin⁡a−asin⁡a−asin⁡x+asin⁡ax−al=\displaystyle \lim_{x \to a}\dfrac{x\sin a-a\sin a-a\sin x+a\sin a}{x-a}
Taking sina common from the first two terms and a common from the last two terms, we get
l=lim⁡x→a(x−a)sin⁡a−a(sin⁡x−sin⁡a)x−al=\displaystyle \lim_{x \to a}\dfrac{\left( x-a \right)\sin a-a\left( \sin x-\sin a \right)}{x-a}
We know that a−bc=ac−bc\dfrac{a-b}{c}=\dfrac{a}{c}-\dfrac{b}{c}
Using the above identity, we get
l=lim⁡x→asin⁡a−lim⁡x→aa(sin⁡x−sin⁡a)x−al=\displaystyle \lim_{x \to a}\sin a-\displaystyle \lim_{x \to a}\dfrac{a\left( \sin x-\sin a \right)}{x-a}
We know that lim⁡x→af(x)−f(a)x−a=f′(a)\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)-f\left( a \right)}{x-a}=f'\left( a \right)
Put f(x) = sinx, we get
lim⁡x→asin⁡x−sin⁡ax−a=cos⁡a\displaystyle \lim_{x \to a}\dfrac{\sin x-\sin a}{x-a}=\cos a
Hence, we have
l=sin⁡a−acos⁡al=\sin a-a\cos a
Hence, we have
lim⁡x→axsin⁡a−asin⁡xx−a=sin⁡a−acos⁡a\displaystyle \lim_{x \to a}\dfrac{x\sin a-a\sin x}{x-a}=\sin a-a\cos a

Note : [1] Alternative Solution: Using L Hospital Rule
Value of numerator at x = a is asin⁡a−asin⁡a=0a\sin a-a\sin a=0
Value of denominator at x = a is a−a=0a-a=0
Hence the form of the given limit is 00\dfrac{0}{0}
We know that if lim⁡x→af(x)g(x)\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)} is form 00\dfrac{0}{0} or ∞∞\dfrac{\infty }{\infty } then, lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x)\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}=\displaystyle \lim_{x \to a}\dfrac{f'\left( x \right)}{g'\left( x \right)} provided the derivatives exist. This is known as L Hospital Rule.
Using L Hospital rule, we get
lim⁡x→axsin⁡a−asin⁡xx−a=lim⁡x→ad(xsin⁡a−asin⁡x)dxd(x−a)dx =lim⁡x→asin⁡a−acos⁡x1=sin⁡a−acos⁡a \begin{aligned} & \displaystyle \lim_{x \to a}\dfrac{x\sin a-a\sin x}{x-a}=\displaystyle \lim_{x \to a}\dfrac{\dfrac{d\left( x\sin a-a\sin x \right)}{dx}}{\dfrac{d\left( x-a \right)}{dx}} \\\ & =\displaystyle \lim_{x \to a}\dfrac{\sin a-a\cos x}{1}=\sin a-a\cos a \\\ \end{aligned}
Which is the same as obtained above.
[2] One should always practice solving limits without using L Hospital rule and series expansion as this helps in understanding the ways to simplify complicated examples as done above.