Question

Evaluate the limit $\displaystyle \lim_{x \to 0}\dfrac{a\sin ax-b\sin bx}{\tan ax-\tan bx}.$
A) a + b
B) a - b
C) 1
D) $\dfrac{{{a}^{2}}}{{{b}^{2}}}$

Answer 1

Hint: In this we compute limit by using algebra of limit i.e 1) limit of sum (difference) of two function is equal to sum (difference) of limit of two function, 2) limit of product of functions is equal to product of functions and 3) limit of ratio of two functions is equal to ratio of limit of functions. By using these properties of limit we first use that limit of ratio is equal to ratio of limit of functions then we will use limit of difference of two functions is equal to difference of limit of function. Then we will use the limit of sinx and tanx as x tends to 0.

Complete step-by-step answer:
Let $\text{L}=\displaystyle \lim_{x \to 0}\dfrac{a\sin ax-b\sin bx}{\tan ax-\tan bx}.$
Now we will use the property that the limit of ratio of the functions is equal to ratio of limit of function.
$\text{L}=\dfrac{\displaystyle \lim_{x \to 0}\left( a\sin ax-b\sin bx \right)}{\displaystyle \lim_{x \to 0}\left( \tan ax-\tan bx \right)}.$
Now we will use the property that the limit of difference of the functions is equal to difference of limit of function to the numerator and denominator.
$\text{L}=\dfrac{\displaystyle \lim_{x \to 0}a\sin ax-\displaystyle \lim_{x \to 0}b\sin bx}{\displaystyle \lim_{x \to 0}\tan ax-\displaystyle \lim_{x \to 0}\tan bx}.$
Since a and b are constant we can take it common.
$\text{L}=\dfrac{\displaystyle \lim_{x \to 0}{\mathop{a\lim }}\,\sin ax-\displaystyle \lim_{x \to 0}{\mathop{b\lim }}\,\sin bx}{\displaystyle \lim_{x \to 0}\tan ax-\displaystyle \lim_{x \to 0}\tan bx}.$
by dividing both numerator and denominator by x, we get
\[\text{L}=\dfrac{\dfrac{\displaystyle \lim_{x \to 0}{\mathop{a\lim }}\,\sin ax-\displaystyle \lim_{x \to 0}{\mathop{b\lim }}\,\sin bx}{x}}{\dfrac{\displaystyle \lim_{x \to 0}\tan ax-\displaystyle \lim_{x \to 0}\tan bx}{x}}.\]
$\text{L}=\dfrac{\displaystyle \lim_{x \to 0}{\mathop{a\lim }}\,\dfrac{\sin ax}{x}-\displaystyle \lim_{x \to 0}{\mathop{b\lim }}\,\dfrac{\sin bx}{x}}{\displaystyle \lim_{x \to 0}\dfrac{\tan ax}{x}-\displaystyle \lim_{x \to 0}\dfrac{\tan bx}{x}}.$

By multiplying and dividing sinax by a, sinbx by b tanax by a and tanbx by b, we get
$\text{L}=\dfrac{{{a}^{2}}\displaystyle \lim_{x \to 0}\dfrac{\sin ax}{ax}-{{b}^{2}}\displaystyle \lim_{x \to 0}\dfrac{\sin bx}{bx}}{a\displaystyle \lim_{x \to 0}\dfrac{\tan ax}{ax}-b\displaystyle \lim_{x \to 0}\dfrac{\tan bx}{bx}}.$
Since we know that, $\displaystyle \lim_{x \to 0}\dfrac{\sin ax}{ax}=1$ and $\displaystyle \lim_{x \to 0}\dfrac{\tan ax}{ax}=1$
$\text{L}=\dfrac{{{a}^{2}}-{{b}^{2}}}{a-b}.$
Since ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$
\[\text{L}=\dfrac{\left( a-b \right)\left( a+b \right)}{a-b}.\]
\[\text{L}=\left( a+b \right).\]
$\displaystyle \lim_{x \to 0}\dfrac{a\sin ax-b\sin bx}{\tan ax-\tan bx}=a+b$

So, the correct answer is “Option A”.

Note: In this problem one should know what x tends to a means. It means that if x takes values closer and closer to a but not equal to a then we say that x tends to a. Also know that limit of function exists when right hand limit of function is equal to left hand limit is equal to value of function at that point.