# 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.

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

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

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

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

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