Math Doubts

Find $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{1-\cos{6x}}{1-\cos{7x}}$

$\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{1-\cos{6x}}{1-\cos{7x}}$

Check the value of the function as $x$ approaches $0$. It can be done by substituting $x = 0$.

$=\,$ $\dfrac{1-\cos{6(0)}}{1-\cos{7(0)}}$

$=\,$ $\dfrac{1-\cos{0}}{1-\cos{0}}$

According to Trigonometry, the value of $\cos{0}$ is equal to $1$.

$=\,$ $\dfrac{1-1}{1-1}$

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

The trigonometric expression is undefined when $x$ tends to $0$. So, an alternative mathematical approach should be used to solve this limit problem in calculus.

Convert Cosine to sine function

According to cos double angle formula, each cosine function can be converted as sine function.

$1-\cos{2\theta} = 2\sin^2{\theta}$

So, change each cos function as cos of double angle function by a mathematical setting.

$=\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{1-\cos{(2 \times 3x)}}{1-\cos{\Bigg(2 \times \dfrac{7x}{2}\Bigg)}}$

$=\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{1-\cos{2(3x)}}{1-\cos{2\Bigg(\dfrac{7x}{2}\Bigg)}}$

$=\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{2\sin^2{(3x)}}{2\sin^2{\Bigg(\dfrac{7x}{2}\Bigg)}}$

$=\,$ $\require{cancel} \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\cancel{2}\sin^2{(3x)}}{\cancel{2}\sin^2{\Bigg(\dfrac{7x}{2}\Bigg)}}$

$=\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin^2{(3x)}}{\sin^2{\Bigg(\dfrac{7x}{2}\Bigg)}}$

Split the expression as multiplying factors

The sine functions in both numerator and denominator have same exponent. So, there will be no issue to use lim sinx/x as x approaches 0 rule. In order to apply it, separate both sine functions as two multiplying factors.

$=\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \sin^2{(3x)} \times \dfrac{1}{\sin^2{\Bigg(\dfrac{7x}{2}\Bigg)}}$

There are two considerable factors for apply the above limit rule. Each sine function should contain its angle as its denominator. Similarly, the exponent of each sine function is $2$. So, the denominator should contain $2$ as its exponent.

$=\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \sin^2{(3x)} \times \dfrac{1}{\sin^2{\Bigg(\dfrac{7x}{2}\Bigg)}} \times 1$

$=\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \sin^2{(3x)} \times \dfrac{1}{\sin^2{\Bigg(\dfrac{7x}{2}\Bigg)}} \times \dfrac{x^2}{x^2}$

$=\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin^2{(3x)}}{x^2} \times \dfrac{x^2}{\sin^2{\Bigg(\dfrac{7x}{2}\Bigg)}}$

Apply Limit Product Rule

The limit value belongs to both functions. It can be written to both functions separately.

$=\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin^2{(3x)}}{x^2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{x^2}{\sin^2{\Bigg(\dfrac{7x}{2}\Bigg)}}$

Simplify the Expression to apply limit rule

Adjust denominator of each sine function such that the denominator is same as the angle of the associated sine function.

$=\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{9 \times \sin^2{(3x)}}{9 \times x^2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\dfrac{49}{4} \times x^2}{\dfrac{49}{4} \times \sin^2{\Bigg(\dfrac{7x}{2}\Bigg)}}$

$=\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{9 \times \sin^2{(3x)}}{9x^2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\dfrac{49}{4}x^2}{\dfrac{49}{4} \times \sin^2{\Bigg(\dfrac{7x}{2}\Bigg)}}$

$=\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{9 \times \sin^2{(3x)}}{{(3x)}^2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{ {\Bigg(\dfrac{7x}{2}\Bigg)}^2}{\dfrac{49}{4} \times \sin^2{\Bigg(\dfrac{7x}{2}\Bigg)}}$

$=\,$ $9 \times \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin^2{(3x)}}{{(3x)}^2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{4 \times {\Bigg(\dfrac{7x}{2}\Bigg)}^2}{49 \times \sin^2{\Bigg(\dfrac{7x}{2}\Bigg)}}$

$=\,$ $9 \times \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin^2{(3x)}}{{(3x)}^2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{4}{49} \times \dfrac{{\Bigg(\dfrac{7x}{2}\Bigg)}^2}{\sin^2{\Bigg(\dfrac{7x}{2}\Bigg)}}$

$=\,$ $9 \times \dfrac{4}{49} \times \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin^2{(3x)}}{{(3x)}^2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{{\Bigg(\dfrac{7x}{2}\Bigg)}^2}{\sin^2{\Bigg(\dfrac{7x}{2}\Bigg)}}$

$=\,$ $\dfrac{9 \times 4}{49} \times \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin^2{(3x)}}{{(3x)}^2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{{\Bigg(\dfrac{7x}{2}\Bigg)}^2}{\sin^2{\Bigg(\dfrac{7x}{2}\Bigg)}}$

$=\,$ $\dfrac{36}{49} \times \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin^2{(3x)}}{{(3x)}^2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{{\Bigg(\dfrac{7x}{2}\Bigg)}^2}{\sin^2{\Bigg(\dfrac{7x}{2}\Bigg)}}$

$=\,$ $\dfrac{36}{49} \times \displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin^2{(3x)}}{{(3x)}^2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{1}{\dfrac{\sin^2{\Bigg(\dfrac{7x}{2}\Bigg)}}{{\Bigg(\dfrac{7x}{2}\Bigg)}^2}}$

$=\,$ $\dfrac{36}{49} \times \displaystyle \large \lim_{x \,\to\, 0} \normalsize {\Bigg[\dfrac{\sin{3x}}{3x}\Bigg]}^2$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{1}{{\Bigg[\dfrac{\sin{\Bigg(\dfrac{7x}{2}\Bigg)}}{\dfrac{7x}{2}}\Bigg]}^2}$

$=\,$ $\dfrac{36}{49} \times {\Bigg[\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin{3x}}{3x}\Bigg]}^2$ $\times$ $\dfrac{1}{{\Bigg[\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin{\Bigg(\dfrac{7x}{2}\Bigg)}}{\dfrac{7x}{2}}\Bigg]}^2}$

Obtain the solution of the problem

Now, set limit value of each function.

If $x \to 0$, then $3x \to 3(0)$. Therefore, $3x \to 0$.

If $x \to 0$ then $\dfrac{7x}{2} \to \dfrac{7}{2} \times 0$. Therefore $\dfrac{7x}{2} \to 0$.

So, limit value of each function can be changed.

$=$ $\dfrac{36}{49} \times {\Bigg[\displaystyle \large \lim_{3x \,\to\, 0} \normalsize \dfrac{\sin{3x}}{3x}\Bigg]}^2$ $\times$ $\dfrac{1}{{\Bigg[\displaystyle \large \lim_{\frac{7x}{2} \,\to\, 0} \normalsize \dfrac{\sin{\Bigg(\dfrac{7x}{2}\Bigg)}}{\dfrac{7x}{2}}\Bigg]}^2}$

Now, each multiplying factor represents lim sinx/x as x approaches 0 rule. So, the value of each multiplying factor is 1.

$=\,$ $\dfrac{36}{49} \times {(1)}^2$ $\times$ $\dfrac{1}{{(1)}^2}$

$=\,$ $\dfrac{36}{49} \times 1$ $\times$ $\dfrac{1}{1}$

$=\,$ $\dfrac{36}{49}\times 1$

$=\,$ $\dfrac{36}{49}$



Follow us
Email subscription
Math Doubts
Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more
Follow us on Social Media
Mobile App for Android users Math Doubts Android App
Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more