Find the solution of \[\dfrac{{sin(ax)}}{{sin(bx)}}\]as x approaches 0.

Hint:First check if the limit is indeterminate form or not. If not, there are different methods to change the indeterminate form to determinate form. In this question you may use L'Hospital#39;s rule. L'Hospital#39;s rule provides a method to judge limits of indeterminate forms.

Complete step by step solution:
If we directly put 0 in place of x, \[\mathop {lim}\limits_{x \to 0} \dfrac{{sin(ax)}}{{sin(bx)}}\]
converts to $\dfrac{0}{0}$, which is an indeterminate form. Hence this method is applicable in this case.

Here we use L'Hospital#39;s rule to change the form of the question.

L’Hospital’s rule states,
\[\mathop {\lim }\limits_{x \to a}
\dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'(x)}}{{g'(x)}}\].

Using the above formula, we can change the original question as,
\[\mathop {lim}\limits_{x \to 0} \dfrac{{sin(ax)}}{{sin(bx)}} = \mathop {lim}\limits_{x \to 0}
\dfrac{{a\;\cos (ax)}}{{b\;\cos (bx)}}\]

Taking the constants out of the limit (as limit doesn’t have any effect on the constants),
\[\mathop {lim}\limits_{x \to 0} \dfrac{{sin(ax)}}{{sin(bx)}} = \dfrac{a}{b}\mathop {lim}\limits_{x
\to 0} \dfrac{{\cos (ax)}}{{\cos (bx)}}\]…………..(i)

We know that, $\cos 0^\circ = 1$

Hence, we can now put x=0 in equation 1,
\[\mathop {lim}\limits_{x \to 0} \dfrac{{sin(ax)}}{{sin(bx)}} = \dfrac{a}{b}\mathop {lim}\limits_{x
\to 0} \dfrac{{\cos (ax)}}{{\cos (bx)}} = \dfrac{a}{b}\mathop {lim}\limits_{x \to 0} \dfrac{1}{1}\;\]

Again, the limit doesn’t affect the constants and hence the final answer is , \[\dfrac{a}{b}\].

Alternate approach:
We know that, \[\dfrac{{\sin \;z}}{z}\]is 1 when z tends to 0, i.e., \[\mathop {lim}\limits_{z \to 0}
\dfrac{{\sin \;z}}{z} = 1\]

We can use this formula to solve the given question.
\[\mathop {lim}\limits_{x \to 0} \dfrac{{sin(ax)}}{{sin(bx)}}\]=
$
= \mathop {lim}\limits_{x \to 0} \dfrac{{\dfrac{{sin(ax)}}{x}}}{{\dfrac{{sin(bx)}}{x}}} \\
= \mathop {lim}\limits_{x \to 0} \dfrac{{\dfrac{{sin(ax)}}{{ax}} \times a}}{{\dfrac{{sin(bx)}}{{bx}}
\times b}} \\
$
\[\mathop { = lim}\limits_{x \to 0}
\dfrac{{\dfrac{{sin(ax)}}{{ax}}}}{{\dfrac{{sin(bx)}}{{bx}}}}\dfrac{a}{b}\]

If we put u=ax and v=bx

Then the above limit will look like,
$\mathop { = lim}\limits_{x \to 0} \dfrac{{\dfrac{{sin(u)}}{u}}}{{\dfrac{{sin(v)}}{v}}}\dfrac{a}{b}$

Now the limit part becomes equal to 1,

Thus the final answer will become $\mathop {lim}\limits_{x
\to 0} \dfrac{1}{1}.\dfrac{a}{b} = \dfrac{a}{b}$

Note: Application of the L'Hopital's rule often converts an indeterminate form to an
expression which will be easily evaluated by substitution. In more mathematical language, if at a given point two functions have an infinite limit or zero as a limit and are both differentiable in an exceedingly neighbourhood of this time then the limit of the quotient of the functions is up to the limit of the quotient of their derivatives given that this limit exists.