Math Doubts

Evaluate $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{1-\cos{mx}}{1-\cos{nx}}}$ by formulas

The limit of the quotient of one minus cosine of angle $mx$ by one minus cos of angle $nx$ as the value of $x$ approaches to zero is indeterminate as per the direct substitution method.

$\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{1-\cos{mx}}{1-\cos{nx}}}$

It clears that the direct substitution is not useful to find the limit. However, it can be evaluated by the combination of both limit rules and other mathematical formulas.

Technique to overcome the indeterminate form

Due to the involvement of the cosine functions in both numerator and denominator, the rational function in trigonometric ratios form becomes indeterminate. Similarly, there is no limit rule in terms of cosine. However, there is a limit rule in sine function in calculus. So, it is recommendable to convert the rational trigonometric function in terms of sine and it is possible by one minus cos double angle identity but there is no double angle in cosine function in the rational trigonometric function.

In order to overcome this issue, let’s try to adjust the angle inside the cosine function as a double angle.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{1-\cos{(1 \times mx)}}{1-\cos{(1 \times nx)}}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{1-\cos{\Big(\dfrac{2}{2} \times mx\Big)}}{1-\cos{\Big(\dfrac{2}{2} \times nx\Big)}}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{1-\cos{\Big(\dfrac{2 \times 1}{2} \times mx\Big)}}{1-\cos{\Big(\dfrac{2 \times 1}{2} \times nx\Big)}}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{1-\cos{\Big(2 \times \dfrac{1}{2} \times mx\Big)}}{1-\cos{\Big(2 \times \dfrac{1}{2} \times nx\Big)}}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{1-\cos{\Big(2 \times \dfrac{1 \times mx}{2}\Big)}}{1-\cos{\Big(2 \times \dfrac{1 \times nx}{2}\Big)}}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{1-\cos{\Big(2 \times \dfrac{mx}{2}\Big)}}{1-\cos{\Big(2 \times \dfrac{nx}{2}\Big)}}}$

The expressions in both numerator and denominator are successfully adjusted to the one minus cosine of double angle identity. Hence, each trigonometric expression can be transformed into the sine squared of angle.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{2\sin^2{\Big(\dfrac{mx}{2}\Big)}}{2\sin^2{\Big(\dfrac{nx}{2}\Big)}}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{2 \times \sin^2{\Big(\dfrac{mx}{2}\Big)}}{2 \times \sin^2{\Big(\dfrac{nx}{2}\Big)}}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\cancel{2} \times \sin^2{\Big(\dfrac{mx}{2}\Big)}}{\cancel{2} \times \sin^2{\Big(\dfrac{nx}{2}\Big)}}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin^2{\Big(\dfrac{mx}{2}\Big)}}{\sin^2{\Big(\dfrac{nx}{2}\Big)}}}$

Preparing the function to find the limit

There is a sine function in both numerator and denominator. The limit will be still indeterminate as the value of $x$ is closer to $0$ as per the direct substitution.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin^2{\Big(\dfrac{mx}{2}\Big) \times 1}}{\sin^2{\Big(\dfrac{nx}{2}\Big)}}}$

However, the result can be changed by adjusting the trigonometric limit function into the trigonometric limit rule in sine function. Now, let us try to transform the function into the trigonometric limit rule by splitting the function.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Bigg(\sin^2{\Big(\dfrac{mx}{2}\Big)} \times \dfrac{1}{\sin^2{\Big(\dfrac{nx}{2}\Big)}}\Bigg)}$

The limit of the product of functions can be calculated by the product of their limits as per the product rule of limits.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \sin^2{\Big(\dfrac{mx}{2}\Big)}}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{1}{\sin^2{\Big(\dfrac{nx}{2}\Big)}}}$

According to the reciprocal limit rule, the limit of reciprocal of a function can be calculated by the reciprocal of its limit.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \sin^2{\Big(\dfrac{mx}{2}\Big)}}$ $\times$ $\dfrac{1}{\displaystyle \large \lim_{x\,\to\,0}{\normalsize \sin^2{\Big(\dfrac{nx}{2}\Big)}}}$

According to the trigonometry, the sine squared of an angle can be written as a square of sine of angle.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\sin{\Big(\dfrac{mx}{2}\Big)}\bigg)^2}$ $\times$ $\dfrac{1}{\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\sin{\Big(\dfrac{nx}{2}\Big)}\bigg)^2}}$

The power rule of limits can be used to the find the limit of a power function in sine by the square of its limit.

$=\,\,\,$ $\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \sin{\Big(\dfrac{mx}{2}\Big)}\bigg)^2}$ $\times$ $\dfrac{1}{\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \sin{\Big(\dfrac{nx}{2}\Big)}\bigg)^2}}$

Now, let’s try to make each function same as the trigonometric limit rule.

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(1 \times \sin{\Big(\dfrac{mx}{2}\Big)}\bigg)\Bigg)^2}$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(1 \times \sin{\Big(\dfrac{nx}{2}\Big)}\bigg)\Bigg)^2}}$

For converting the both factors as a trigonometric limit rule in sine function, the sine functions in first and second factors should have $\dfrac{mx}{2}$ and $\dfrac{nx}{2}$ as their denominators respectively. So, write the number $1$ in first factor in terms of $\dfrac{mx}{2}$ and the number $1$ in second factor in terms of $\dfrac{nx}{2}$.

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{\Big(\dfrac{mx}{2}\Big)}{\Big(\dfrac{mx}{2}\Big)} \times \sin{\Big(\dfrac{mx}{2}\Big)}\bigg)\Bigg)^2}$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{\Big(\dfrac{nx}{2}\Big)}{\Big(\dfrac{nx}{2}\Big)} \times \sin{\Big(\dfrac{nx}{2}\Big)}\bigg)\Bigg)^2}}$

It is time to focus on expressing each factor as a trigonometric limit rule in sine function and use the multiplication technique of the fractions for obtaining it.

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{\Big(\dfrac{mx}{2}\Big) \times 1}{\Big(\dfrac{mx}{2}\Big)} \times \sin{\Big(\dfrac{mx}{2}\Big)}\bigg)\Bigg)^2}$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{\Big(\dfrac{nx}{2}\Big) \times 1}{\Big(\dfrac{nx}{2}\Big)} \times \sin{\Big(\dfrac{nx}{2}\Big)}\bigg)\Bigg)^2}}$

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\Big(\dfrac{mx}{2}\Big) \times \dfrac{1}{\Big(\dfrac{mx}{2}\Big)} \times \sin{\Big(\dfrac{mx}{2}\Big)}\bigg)\Bigg)^2}$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\Big(\dfrac{nx}{2}\Big) \times \dfrac{1}{\Big(\dfrac{nx}{2}\Big)} \times \sin{\Big(\dfrac{nx}{2}\Big)}\bigg)\Bigg)^2}}$

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\Big(\dfrac{mx}{2}\Big) \times \dfrac{1 \times \sin{\Big(\dfrac{mx}{2}\Big)}}{\Big(\dfrac{mx}{2}\Big)}\bigg)\Bigg)^2}$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\Big(\dfrac{nx}{2}\Big) \times \dfrac{1 \times \sin{\Big(\dfrac{nx}{2}\Big)}}{\Big(\dfrac{nx}{2}\Big)}\bigg)\Bigg)^2}}$

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\Big(\dfrac{mx}{2}\Big) \times \dfrac{\sin{\Big(\dfrac{mx}{2}\Big)}}{\Big(\dfrac{mx}{2}\Big)}\bigg)\Bigg)^2}$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\Big(\dfrac{nx}{2}\Big) \times \dfrac{\sin{\Big(\dfrac{nx}{2}\Big)}}{\Big(\dfrac{nx}{2}\Big)}\bigg)\Bigg)^2}}$

According to the product rule of limits, the limit of product of two functions can be evaluated by product of their limits in both factors.

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{mx}{2}\Big)} \times \displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{mx}{2}\Big)}}{\Big(\dfrac{mx}{2}\Big)}\Bigg)^2}$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{nx}{2}\Big)} \times \displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{nx}{2}\Big)}}{\Big(\dfrac{nx}{2}\Big)}\Bigg)^2}}$

The power of a product rule can be used for both factors for splitting the square of two functions as a product of their squares.

$=\,\,\,$ $\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{mx}{2}\Big)\bigg)^2}$ $\times$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{mx}{2}\Big)}}{\Big(\dfrac{mx}{2}\Big)}\Bigg)^2}$ $\times$ $\dfrac{1}{\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{nx}{2}\Big)\bigg)^2} \times \Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{nx}{2}\Big)}}{\Big(\dfrac{nx}{2}\Big)}}\Bigg)^2}$

One factor is split as a product of two factors but the second factor is not split as a product of two factors completely and it is now third factor. So, let’s focus on splitting the third factor as per the multiplication rule of the fractions and it helps us in finding the limit of the given trigonometric rational function.

$=\,\,\,$ $\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{mx}{2}\Big)\bigg)^2}$ $\times$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{mx}{2}\Big)}}{\Big(\dfrac{mx}{2}\Big)}\Bigg)^2}$ $\times$ $\dfrac{1 \times 1}{\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{nx}{2}\Big)\bigg)^2} \times \Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{nx}{2}\Big)}}{\Big(\dfrac{nx}{2}\Big)}}\Bigg)^2}$

$=\,\,\,$ $\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{mx}{2}\Big)\bigg)^2}$ $\times$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{mx}{2}\Big)}}{\Big(\dfrac{mx}{2}\Big)}\Bigg)^2}$ $\times$ $\dfrac{1}{\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{nx}{2}\Big)\bigg)^2} }$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{nx}{2}\Big)}}{\Big(\dfrac{nx}{2}\Big)}}\Bigg)^2}$

For our convenience, write the algebraic limit functions closer by separating the trigonometric limit functions.

$=\,\,\,$ $\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{mx}{2}\Big)\bigg)^2}$ $\times$ $\dfrac{1}{\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{nx}{2}\Big)\bigg)^2} }$ $\times$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{mx}{2}\Big)}}{\Big(\dfrac{mx}{2}\Big)}\Bigg)^2}$ $\times$ $\dfrac{1}{\displaystyle \large \Bigg(\lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{nx}{2}\Big)}}{\Big(\dfrac{nx}{2}\Big)}}\Bigg)^2}$

Now, focus on first two factors and they both are algebraic limit functions. So, let’s us try to simplify by multiplying them as per the multiplication property of fractions.

$=\,\,\,$ $\dfrac{\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{mx}{2}\Big)\bigg)^2} \times 1}{\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{nx}{2}\Big)\bigg)^2} }$ $\times$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{mx}{2}\Big)}}{\Big(\dfrac{mx}{2}\Big)}\Bigg)^2}$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{nx}{2}\Big)}}{\Big(\dfrac{nx}{2}\Big)}}\Bigg)^2}$

$=\,\,\,$ $\dfrac{\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{mx}{2}\Big)\bigg)^2}}{\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{nx}{2}\Big)\bigg)^2} }$ $\times$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{mx}{2}\Big)}}{\Big(\dfrac{mx}{2}\Big)}\Bigg)^2}$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{nx}{2}\Big)}}{\Big(\dfrac{nx}{2}\Big)}}\Bigg)^2}$

The quotient of squares of two functions can be simplified by the square of their quotient as per the power of a quotient rule.

$=\,\,\,$ $\Bigg(\dfrac{\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{mx}{2}\Big)}}{\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{nx}{2}\Big)}}\Bigg)^2$ $\times$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{mx}{2}\Big)}}{\Big(\dfrac{mx}{2}\Big)}\Bigg)^2}$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{nx}{2}\Big)}}{\Big(\dfrac{nx}{2}\Big)}}\Bigg)^2}$

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\Big(\dfrac{mx}{2}\Big)}{\Big(\dfrac{nx}{2}\Big)}\Bigg)^2}$ $\times$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{mx}{2}\Big)}}{\Big(\dfrac{mx}{2}\Big)}\Bigg)^2}$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{nx}{2}\Big)}}{\Big(\dfrac{nx}{2}\Big)}}\Bigg)^2}$

Let us simplify the mathematical function in the first factor.

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\Big(\dfrac{mx}{2}\Big) \times 1}{\Big(\dfrac{nx}{2}\Big)}\Bigg)^2}$ $\times$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{mx}{2}\Big)}}{\Big(\dfrac{mx}{2}\Big)}\Bigg)^2}$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{nx}{2}\Big)}}{\Big(\dfrac{nx}{2}\Big)}}\Bigg)^2}$

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\Big(\dfrac{mx}{2}\Big) \times \dfrac{1}{\Big(\dfrac{nx}{2}\Big)}\bigg)\Bigg)^2}$ $\times$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{mx}{2}\Big)}}{\Big(\dfrac{mx}{2}\Big)}\Bigg)^2}$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{nx}{2}\Big)}}{\Big(\dfrac{nx}{2}\Big)}}\Bigg)^2}$

$=\,\,\,$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{mx}{2} \times \dfrac{1}{\dfrac{nx}{2}}\bigg)\Bigg)^2}$ $\times$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{mx}{2}\Big)}}{\Big(\dfrac{mx}{2}\Big)}\Bigg)^2}$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{nx}{2}\Big)}}{\Big(\dfrac{nx}{2}\Big)}}\Bigg)^2}$

$=\,\,\,$ $\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\dfrac{mx}{2} \times \dfrac{2}{nx}\Big)\bigg)^2}$ $\times$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{mx}{2}\Big)}}{\Big(\dfrac{mx}{2}\Big)}\Bigg)^2}$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{nx}{2}\Big)}}{\Big(\dfrac{nx}{2}\Big)}}\Bigg)^2}$

$=\,\,\,$ $\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{mx \times 2}{2 \times nx}\bigg)^2}$ $\times$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{mx}{2}\Big)}}{\Big(\dfrac{mx}{2}\Big)}\Bigg)^2}$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{nx}{2}\Big)}}{\Big(\dfrac{nx}{2}\Big)}}\Bigg)^2}$

$=\,\,\,$ $\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{m \times x \times 2}{2 \times n \times x}\bigg)^2}$ $\times$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{mx}{2}\Big)}}{\Big(\dfrac{mx}{2}\Big)}\Bigg)^2}$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{nx}{2}\Big)}}{\Big(\dfrac{nx}{2}\Big)}}\Bigg)^2}$

$=\,\,\,$ $\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{m \times \cancel{x} \times \cancel{2}}{\cancel{2} \times n \times \cancel{x}}\bigg)^2}$ $\times$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{mx}{2}\Big)}}{\Big(\dfrac{mx}{2}\Big)}\Bigg)^2}$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{nx}{2}\Big)}}{\Big(\dfrac{nx}{2}\Big)}}\Bigg)^2}$

Find the Limit of the function

The limit of a trigonometric function is split as a product of limits of three functions. So, let’s find the limit of each function for finding the limit of the given trigonometric function.

$=\,\,\,$ $\bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{m}{n}\bigg)^2}$ $\times$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{mx}{2}\Big)}}{\Big(\dfrac{mx}{2}\Big)}\Bigg)^2}$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{nx}{2}\Big)}}{\Big(\dfrac{nx}{2}\Big)}}\Bigg)^2}$

Use the direct substitution method to find the limit of the function in the first factor as the value of $x$ approaches to zero.

$=\,\,\,$ $\Big(\dfrac{m}{n}\Big)^2$ $\times$ $\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{mx}{2}\Big)}}{\Big(\dfrac{mx}{2}\Big)}\Bigg)^2}$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{nx}{2}\Big)}}{\Big(\dfrac{nx}{2}\Big)}}\Bigg)^2}$

It is time to focus on second and third factors. The functions in both factors are similar to the trigonometric limit rule in sine function.

$(1).\,\,\,$ If $x \,\to\, 0$ then $\dfrac{m}{2} \times x \,\to\,\dfrac{m}{2} \times 0$. Therefore, $\dfrac{mx}{2} \,\to\,0$

$(2).\,\,\,$ if $x \,\to\, 0$ then $\dfrac{n}{2} \times x \,\to\,\dfrac{n}{2} \times 0$. Therefore, $\dfrac{nx}{2} \,\to\,0$

We have proved that when $x$ approaches to $0$, the value of $\dfrac{mx}{2}$ and $\dfrac{nx}{2}$ are also closer to zero.

$=\,\,\,$ $\Big(\dfrac{m}{n}\Big)^2$ $\times$ $\Bigg(\displaystyle \large \lim_{\normalsize \dfrac{mx}{2}\large\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{mx}{2}\Big)}}{\Big(\dfrac{mx}{2}\Big)}\Bigg)^2}$ $\times$ $\dfrac{1}{\Bigg(\displaystyle \large \lim_{\normalsize \dfrac{nx}{2}\large\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{nx}{2}\Big)}}{\Big(\dfrac{nx}{2}\Big)}}\Bigg)^2}$

Suppose, $y \,=\, \dfrac{mx}{2}$ and $z \,=\, \dfrac{nx}{2}$ then express second and third factors in terms of $y$ and $z$ respectively.

$=\,\,\,$ $\Big(\dfrac{m}{n}\Big)^2$ $\times$ $\bigg(\displaystyle \large \lim_{y\,\to\,0}{\normalsize \dfrac{\sin{(y)}}{(y)}\bigg)^2}$ $\times$ $\dfrac{1}{\bigg(\displaystyle \large \lim_{\normalsize z\,\to\,0}{\normalsize \dfrac{\sin{(z)}}{(z)}}\bigg)^2}$

$=\,\,\,$ $\Big(\dfrac{m}{n}\Big)^2$ $\times$ $\bigg(\displaystyle \large \lim_{y\,\to\,0}{\normalsize \dfrac{\sin{y}}{y}\bigg)^2}$ $\times$ $\dfrac{1}{\bigg(\displaystyle \large \lim_{\normalsize z\,\to\,0}{\normalsize \dfrac{\sin{z}}{z}}\bigg)^2}$

According to the trigonometric limit rule in sine function, the limit of sin of angle $y$ by $y$ as $y$ approaches $0$ and the limit of sine of $z$ by $z$ as $z$ approaches $0$ are one.

$=\,\,\,$ $\Big(\dfrac{m}{n}\Big)^2$ $\times$ $(1)^2$ $\times$ $\dfrac{1}{(1)^2}$

It is time to simplify the mathematical expression by multiplying the factors.

$=\,\,\,$ $\Big(\dfrac{m}{n}\Big)^2$ $\times$ $1$ $\times$ $\dfrac{1}{1}$

$=\,\,\,$ $\Big(\dfrac{m}{n}\Big)^2$ $\times$ $1$ $\times$ $1$

$=\,\,\,$ $\Big(\dfrac{m}{n}\Big)^2$ $\times$ $1$

$=\,\,\,$ $\Big(\dfrac{m}{n}\Big)^2$

The square of the quotient of $m$ by $n$ can also be written as the quotient of their squares.

$=\,\,\,$ $\dfrac{m^2}{n^2}$

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