Question

Find the value of the limit $\displaystyle \lim_{x \to a}\dfrac{x\sin a-a\sin x}{x-a}$.

Answer 1

Hint: Add and subtract asina in the numerator. Hence prove that the given limit is equal to $\displaystyle \lim_{x \to a}\dfrac{\left( x-a \right)\sin a}{x-a}+\dfrac{a\left( \sin a-\sin x \right)}{x-a}$.
Use the fact that $\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)-f\left( a \right)}{x-a}=f'\left( a \right)$. 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
$l=\displaystyle \lim_{x \to a}\dfrac{x\sin a-a\sin x}{x-a}$
Adding and subtracting asina in the numerator of RHS, we get
$l=\displaystyle \lim_{x \to a}\dfrac{x\sin a-a\sin x-a\sin a+a\sin a}{x-a}$
Re-writing the terms in the numerator in different order, we get
$l=\displaystyle \lim_{x \to a}\dfrac{x\sin a-a\sin a-a\sin x+a\sin a}{x-a}$
Taking sina common from the first two terms and a common from the last two terms, we get
$l=\displaystyle \lim_{x \to a}\dfrac{\left( x-a \right)\sin a-a\left( \sin x-\sin a \right)}{x-a}$
We know that \[\dfrac{a-b}{c}=\dfrac{a}{c}-\dfrac{b}{c}\]
Using the above identity, we get
$l=\displaystyle \lim_{x \to a}\sin a-\displaystyle \lim_{x \to a}\dfrac{a\left( \sin x-\sin a \right)}{x-a}$
We know that $\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)-f\left( a \right)}{x-a}=f'\left( a \right)$
Put f(x) = sinx, we get
$\displaystyle \lim_{x \to a}\dfrac{\sin x-\sin a}{x-a}=\cos a$
Hence, we have
$l=\sin a-a\cos a$
Hence, we have
$\displaystyle \lim_{x \to a}\dfrac{x\sin a-a\sin x}{x-a}=\sin a-a\cos a$

Note: [1] Alternative Solution: Using L Hospital Rule
Value of numerator at x = a is $a\sin a-a\sin a=0$
Value of denominator at x = a is $a-a=0$
Hence the form of the given limit is $\dfrac{0}{0}$
We know that if $\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}$ is form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$ then, $\displaystyle \lim_{x \to a}\dfrac{f\left( x \right)}{g\left( x \right)}=\displaystyle \lim_{x \to a}\dfrac{f'\left( x \right)}{g'\left( x \right)}$ provided the derivatives exist. This is known as L Hospital Rule.
Using L Hospital rule, we get
$\begin{align}
  & \displaystyle \lim_{x \to a}\dfrac{x\sin a-a\sin x}{x-a}=\displaystyle \lim_{x \to a}\dfrac{\dfrac{d\left( x\sin a-a\sin x \right)}{dx}}{\dfrac{d\left( x-a \right)}{dx}} \\
 & =\displaystyle \lim_{x \to a}\dfrac{\sin a-a\cos x}{1}=\sin a-a\cos a \\
\end{align}$
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.