Question

Evaluate the following:
$\displaystyle \lim_{x \to 0}\dfrac{\sin ax}{\sin bx}$ where $a,b\ne 0$ $$$$

Answer 1

We multiply $x$ in the numerator and denominator of the quotient function $\dfrac{\sin ax}{\sin bx}$ . Then we divide and multiply $a$ with the numerator$\sin ax$. We also divide and multiply $b$ with the denominator$\sin bx$. We use the law of multiplication and division of limits and also the standard limit $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1$ to evaluate.

Complete step by step answer:
We know that limiting value for any real valued single variable function $f\left( x \right)$ when the variable $x$ approaches to real number $a$ in the domain $f\left( x \right)$ is denoted by
$\displaystyle \lim_{x \to a}f\left( x \right)=L$
Here $L$ is called the limit of the function.
The limit $L$ exists for real valued single variable function $f\left( x \right)$ at any point $x=a$ then if and only if Left hand limit(LHL)= right hand limit(RHL)=the value of the function at $x=a$. In symbols,

& \text{LHL}=\text{RHL=}f\left( a \right) \\\ & \Rightarrow \underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right) \\\ \end{aligned}$$ The multiplication law of limits is given by for two functions $f\left( x \right)$ and $g\left( x \right)$ $$\displaystyle \lim_{x \to a}f\left( x \right)\cdot \displaystyle \lim_{x \to a}g\left( x \right)=\displaystyle \lim_{x \to a}\left( f\left( x \right)\cdot g\left( x \right) \right)$$ The division rule of limits for two functions $f\left( x \right)$ and $g\left( x \right)\ne 0$ is, $$\dfrac{\displaystyle \lim_{x \to a}f\left( x \right)}{\displaystyle \lim_{x \to a}g\left( x \right)}=\displaystyle \lim_{x \to a}\left( \dfrac{f\left( x \right)}{g\left( x \right)} \right)$$ We also know that the limit of a constant is constant $$\displaystyle \lim_{x \to a}C=C$$ We know the standard limit for sine function $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1$. We The given function to evaluate limit is $$\displaystyle \lim_{x \to 0}\dfrac{\sin ax}{\sin bx}\text{ where }a,b\ne 0$$ Let us multiply $x$ in the numerator and denominator. $$\begin{aligned} & \displaystyle \lim_{x \to 0}\dfrac{\sin ax}{x}\text{ }\times \dfrac{x}{\sin bx} \\\ & =\displaystyle \lim_{x \to 0}\dfrac{\dfrac{\sin ax}{x}}{\dfrac{\sin bx}{x}}\text{ } \\\ \end{aligned}$$ Let us multiply and divide $a$ in the numerator and also multiply and divide $b$ with in the denominator. We can do that because we have been provided with the condition $a\ne 0,b\ne 0$.So we have $$=\displaystyle \lim_{x \to 0}\dfrac{\dfrac{\sin ax}{ax}\times a}{\dfrac{\sin bx}{bx}\times b}\text{ }$$ Now we use law of division and then law of multiplication to get , $$\begin{aligned} & =\displaystyle \lim_{x \to 0}\dfrac{\dfrac{\sin ax}{ax}\times a}{\dfrac{\sin bx}{bx}\times b}\text{ } \\\ & \text{=}\dfrac{\displaystyle \lim_{x \to 0}\dfrac{\sin ax}{ax}\times a}{\displaystyle \lim_{x \to 0}\dfrac{\sin bx}{bx}\times b}\text{ } \\\ & \text{=}\dfrac{\displaystyle \lim_{x \to 0}\dfrac{\sin ax}{ax}\times \displaystyle \lim_{x \to 0}a}{\displaystyle \lim_{x \to 0}\dfrac{\sin bx}{bx}\times \displaystyle \lim_{x \to 0}b}\text{ } \\\ \end{aligned}$$ We know that the limit of constant is constant . So we have, $$\begin{aligned} & \text{ =}\dfrac{\displaystyle \lim_{x \to 0}\dfrac{\sin ax}{ax}\times \displaystyle \lim_{x \to 0}a}{\displaystyle \lim_{x \to 0}\dfrac{\sin bx}{bx}\times \displaystyle \lim_{x \to 0}b}\text{ } \\\ & \text{=}\dfrac{\underset{x\to 0}{\mathop{a\lim }}\,\dfrac{\sin ax}{ax}}{\underset{x\to 0}{\mathop{b\lim }}\,\dfrac{\sin bx}{bx}}\text{ } \\\ \end{aligned}$$ Now we substitute $ax=u,bx=v$ in the above result to get, $$\text{ =}\dfrac{\underset{x\to 0}{\mathop{a\lim }}\,\dfrac{\sin u}{u}}{\underset{x\to 0}{\mathop{b\lim }}\,\dfrac{\sin v}{v}}\text{ }$$ Now we use the standard limit $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1$ and get, $$\text{ =}\dfrac{\underset{x\to 0}{\mathop{a\lim }}\,\dfrac{\sin u}{u}}{\underset{x\to 0}{\mathop{b\lim }}\,\dfrac{\sin v}{v}}\text{ =}\dfrac{a\times 1}{b\times 1}=\dfrac{a}{b}$$ **So the value of the limit is $\dfrac{a}{b}$** **Note:** We can alternatively solve the problem with L'Hopital's rule for the indeterminate form $\dfrac{0}{0}$. Here we differentiate the numerator and the denominator until the limiting value at the numerator is non-zero.