The limit of one minus cos of mx by one minus cos of nx as x approaches 0 is indeterminate as per the direct substitution but we have evaluated the limit of this trigonometric rational function fundamentally by the formulas. Due to the indeterminate form, the limit of one minus $\cos{mx}$ by one minus $\cos{nx}$ as $x$ is closer to $0$ can also be evaluated in another method.

It is time to apply the L’Hopital’s law to find the limit by differentiating the expressions in both numerator and denominator with respect to $x$.

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

The derivative of the difference of the functions in both numerator and denominator can be calculated by difference of their derivatives as per the subtraction property of the differentiation.

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

According to the constant derivative property, the differentiation of one with respect to one is equal to zero.

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

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

In calculus, there are no formulas to find the derivatives of cosines of angle $mx$ and $nx$ but there is a formula to calculate the differentiation of cosine of angle $x$. The $\cos{mx}$ and $\cos{nx}$ both are composite functions. Hence, the derivatives of both $\cos{mx}$ and $\cos{nx}$ can be calculated with respect to $x$ as per the chain rule.

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

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

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{mx} \times \dfrac{d}{dx}\,(m \times x)}{\sin{nx} \times \dfrac{d}{dx}\,(n \times x)}}$

The literal coefficients of $x$ in both numerator and denominator can be ignored while differentiating them as per the constant multiple rule of differentiation.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{mx} \times m \times \dfrac{d}{dx}\,(x)}{\sin{nx} \times n \times \dfrac{d}{dx}\,(x)}}$

Use the derivative rule of a variable to find the derivative of $x$ with respect to $x$ in both numerator and denominator of this rational trigonometric function.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{mx} \times m \times 1}{\sin{nx} \times n \times 1}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{mx} \times m}{\sin{nx} \times n}}$

$\therefore\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{m \times \sin{mx}}{n \times \sin{nx}}}$

The L’Hospital’s rule is applied successfully and it is time to calculate the limit of the function as the value of $x$ is closer to $0$ by the direct substitution.

$=\,\,\,$ $\dfrac{m \times \sin{m (0)}}{n \times \sin{n (0)}}$

$=\,\,\,$ $\dfrac{m \times \sin{(m \times 0)}}{n \times \sin{(n \times 0)}}$

$=\,\,\,$ $\dfrac{m \times \sin{(0)}}{n \times \sin{(0)}}$

According to the trigonometry, the sine of zero radian is equal to zero.

$=\,\,\,$ $\dfrac{m \times 0}{n \times 0}$

$=\,\,\,$ $\dfrac{0}{0}$

It is indeterminate and it clears that the first trial of L’Hopital’s Rule is not successful.

Using the L’Hospital’s rule once is not enough to find the limit of the trigonometric rational function and it clears that the L’Hopital’s property should be applied to the function one more time to obtain the limit. So, let’s apply the L’Hospital’s law to the function one more time.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\dfrac{d}{dx}(m \times \sin{mx})}{\dfrac{d}{dx}(n \times \sin{nx})}}$

The literals $m$ and $n$ represent two constants in this rational function and they can be excluded from the differentiation as per the constant multiple rule of the derivatives.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{m \times \dfrac{d}{dx}(\sin{mx})}{n \times \dfrac{d}{dx}(\sin{nx})}}$

There is a derivative rule to find the differentiation of the sine function but $\sin{mx}$ and $\sin{nx}$ both are composite functions. Hence, the derivative rule of sine function cannot be applied to them directly. However, it can be applied to both functions as per the chain rule.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{m \times \cos{mx} \times \dfrac{d}{dx}(mx)}{n \times \cos{nx} \times \dfrac{d}{dx}(nx)}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{m \times \cos{mx} \times \dfrac{d}{dx}(m \times x)}{n \times \cos{nx} \times \dfrac{d}{dx}(n \times x)}}$

The constant factor can be excluded from the differentiation by the constant multiple rule of derivatives.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{m \times \cos{mx} \times m \times \dfrac{d}{dx}(x)}{n \times \cos{nx} \times n \times \dfrac{d}{dx}(x)}}$

Now, use the differentiation rule of a variable to find the derivative of variable $x$ with respect to $x$.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{m \times \cos{mx} \times m \times 1}{n \times \cos{nx} \times n \times 1}}$

It is time to simplify the expressions in both numerator and denominator in the rational trigonometric function.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{m \times \cos{mx} \times m}{n \times \cos{nx} \times n}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{m \times m \times \cos{mx}}{n \times n \times \cos{nx}}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{m^2 \times \cos{mx}}{n^2 \times \cos{nx}}}$

Let’s try to find the limit of the rational function by the direct substitution as the value of $x$ tends to $0$.

$=\,\,\,$ $\dfrac{m^2 \times \cos{(m(0))}}{n^2 \times \cos{(n(0))}}$

$=\,\,\,$ $\dfrac{m^2 \times \cos{(m \times 0)}}{n^2 \times \cos{(n \times 0)}}$

$=\,\,\,$ $\dfrac{m^2 \times \cos{(0)}}{n^2 \times \cos{(0)}}$

According to the trigonometric ratios, the value of cosine of angle zero radian is equal to one.

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

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

Latest Math Topics

Jul 20, 2023

Jun 26, 2023

Jun 23, 2023

Latest Math Problems

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved