Evaluate $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1-\cos{x}\sqrt{\cos{2x}}}{x^2}}$

$x$ is a variable, $\cos{x}$ and $\cos{2x}$ are two trigonometric functions. The functions $\cos{x}$, square root of $\cos{2x}$ and $x$ squared functions formed a special algebraic trigonometric function in a fraction form.

Evaluate the Limit by Direct substitution

Firstly, try direct substitution method to evaluate the limit of this algebraic trigonometric function as $x$ approaches zero.

$= \,\,\,$ $\dfrac{1-\cos{(0)}\sqrt{\cos{2(0)}}}{{(0)}^2}$

$= \,\,\,$ $\dfrac{1-1 \times \sqrt{\cos{(0)}}}{0}$

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

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

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

The limit of the function as $x$ tends to $0$ is indeterminate. So, it should be calculated in another way in calculus.

Evaluate the Limit by Rationalization

Now, rationalize the numerator for releasing the numerator from the radical form.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1-\cos{x}\sqrt{\cos{2x}}}{x^2} \times 1}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[\dfrac{1-\cos{x}\sqrt{\cos{2x}}}{x^2}}$ $\times$ $\dfrac{1+\cos{x}\sqrt{\cos{2x}}}{1+\cos{x}\sqrt{\cos{2x}}}\Bigg]$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[\dfrac{(1-\cos{x}\sqrt{\cos{2x}}) \times (1+\cos{x}\sqrt{\cos{2x}})}{x^2 \times (1+\cos{x}\sqrt{\cos{2x}})}\Bigg]}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[\dfrac{(1-\cos{x}\sqrt{\cos{2x}}) \times (1+\cos{x}\sqrt{\cos{2x}})}{x^2(1+\cos{x}\sqrt{\cos{2x}})}\Bigg]}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[\dfrac{{(1)}^2-{(\cos{x}\sqrt{\cos{2x}})}^2}{x^2(1+\cos{x}\sqrt{\cos{2x}})}\Bigg]}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[\dfrac{1-{(\cos{x})}^2 \times {(\sqrt{\cos{2x}})}^2}{x^2(1+\cos{x}\sqrt{\cos{2x}})}\Bigg]}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[\dfrac{1-\cos^2{x} \times \cos{2x}}{x^2(1+\cos{x}\sqrt{\cos{2x}})}\Bigg]}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[\dfrac{1-\cos^2{x}\cos{2x}}{x^2(1+\cos{x}\sqrt{\cos{2x}})}\Bigg]}$

Methods of Evaluating the Limit

There are two terms in the numerator where $\cos{2x}$ is a factor in the second term of the trigonometric expression and it should be expanded for simplifying the trigonometric expression in the numerator. There are two ways to expand the cos double angle function as per cos double angle identity. You can follow any one of the following methods to evaluate the limit of this trigonometric function.

Method: 1

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1-\cos^2{x}\cos{2x}}{x^2(1+\cos{x}\sqrt{\cos{2x}})}}$

Expand cos functions of numerator in terms of sine

In this method, $\cos^2{x}$ and $\cos{2x}$ are expanded in terms of sin function as per Pythagorean identity and cosine double angle formulas respectively.

$(1) \,\,\,$ $\cos^2{x} \,=\, 1-\sin^2{x}$

$(2) \,\,\,$ $\cos{2x} \,=\, 1-2\sin^2{x}$

Now, replace the $\cos^2{x}$ and $\cos{2x}$ functions in the numerator to simplify the trigonometric expression in the numerator further.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1-(1-\sin^2{x})(1-2\sin^2{x})}{x^2(1+\cos{x}\sqrt{\cos{2x}})}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1-(1-2\sin^2{x}-\sin^2{x}+2\sin^4{x})}{x^2(1+\cos{x}\sqrt{\cos{2x}})}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1-(1-3\sin^2{x}+2\sin^4{x})}{x^2(1+\cos{x}\sqrt{\cos{2x}})}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1-1+3\sin^2{x}-2\sin^4{x}}{x^2(1+\cos{x}\sqrt{\cos{2x}})}}$

$= \,\,\,$ $\require{cancel} \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\cancel{1}-\cancel{1}+3\sin^2{x}-2\sin^4{x}}{x^2(1+\cos{x}\sqrt{\cos{2x}})}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{3\sin^2{x}-2\sin^4{x}}{x^2(1+\cos{x}\sqrt{\cos{2x}})}}$

Factorize the function

Now, split the function as the following two factors.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[\dfrac{3\sin^2{x}-2\sin^4{x}}{x^2} \times \dfrac{1}{1+\cos{x}\sqrt{\cos{2x}}}\Bigg]}$

Now, use the product rule of limits to get the limits of the multiplying functions by the product of their limits.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{3\sin^2{x}-2\sin^4{x}}{x^2}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{1+\cos{x}\sqrt{\cos{2x}}}}$

Use Direct substitution method

Find the limit of the second multiplying function by using direct substitution method.

$= \,\,\,$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{3\sin^2{x}-2\sin^4{x}}{x^2}}\Bigg)$ $\times$ $\Bigg(\dfrac{1}{1+\cos{(0)}\sqrt{\cos{2(0)}}}\Bigg)$

$= \,\,\,$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{3\sin^2{x}-2\sin^4{x}}{x^2}}\Bigg)$ $\times$ $\Bigg(\dfrac{1}{1+1 \times \sqrt{\cos{(0)}}}\Bigg)$

$= \,\,\,$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{3\sin^2{x}-2\sin^4{x}}{x^2}}\Bigg)$ $\times$ $\Bigg(\dfrac{1}{1+\sqrt{1}}\Bigg)$

$= \,\,\,$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{3\sin^2{x}-2\sin^4{x}}{x^2}}\Bigg)$ $\times$ $\Bigg(\dfrac{1}{1+1}\Bigg)$

$= \,\,\,$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{3\sin^2{x}-2\sin^4{x}}{x^2}}\Bigg)$ $\times$ $\Bigg(\dfrac{1}{2}\Bigg)$

$= \,\,\,$ $\Bigg(\dfrac{1}{2}\Bigg)$ $\times$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{3\sin^2{x}-2\sin^4{x}}{x^2}}\Bigg)$

$= \,\,\,$ $\dfrac{1}{2} \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{3\sin^2{x}-2\sin^4{x}}{x^2}}\Bigg)$

Simplify the Algebraic trigonometric function

$\sin^2{x}$ is a common factor in the each term of the trigonometric expression in the numerator. So, take it common from them for simplifying this trigonometric expression further.

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

Now, factorise this algebraic trigonometric function as follows.

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

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

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

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

Evaluate the Limits of the both functions

The limit of the sinx/x as x approaches zero is equal to one.

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

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

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

Now, find the limit of the function by the direct substitution method.

$= \,\,\,$ $\dfrac{1}{2} \Big(3-2\sin^2{(0)}\Big)$

$= \,\,\,$ $\dfrac{1}{2} \Big(3-2 \times {(0)}^2\Big)$

$= \,\,\,$ $\dfrac{1}{2} \Big(3-2 \times 0\Big)$

$= \,\,\,$ $\dfrac{1}{2} \Big(3-0\Big)$

$= \,\,\,$ $\dfrac{1}{2} \Big(3\Big)$

$= \,\,\,$ $\dfrac{1 \times 3}{2}$

$\therefore \,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1-\cos{x}\sqrt{\cos{2x}}}{x^2}}$ $\,=\,$ $\dfrac{3}{2}$

Therefore, it is evaluated that the limit of the quotient of the $1-\cos{x}\sqrt{\cos{2x}}$ by $x^2$ as $x$ approaches zero is equal to $\dfrac{3}{2}$.

Method: 2

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1-\cos^2{x}\cos{2x}}{x^2(1+\cos{x}\sqrt{\cos{2x}})}}$

Expand cos functions of numerator in terms of cosine

In this method, the $\cos{2x}$ function is expanded in terms of cosine as per cos double angle identity.

$\cos{2x} \,=\, 2\cos^2{x}-1$

Now, substitute the $\cos{2x}$ by its equivalent value in the numerator of the function.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1-\cos^2{x}(2\cos^2{x}-1)}{x^2(1+\cos{x}\sqrt{\cos{2x}})}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1-2\cos^4{x}+\cos^2{x}}{x^2(1+\cos{x}\sqrt{\cos{2x}})}}$

Factorize the trigonometric expression

The trigonometric expression in the numerator of this function is a quadratic equation and it can be factored by the factorization.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{-(2\cos^4{x}-\cos^2{x}-1)}{x^2(1+\cos{x}\sqrt{\cos{2x}})}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{-(2\cos^4{x}-2\cos^2{x}+\cos^2{x}-1)}{x^2(1+\cos{x}\sqrt{\cos{2x}})}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{-(2\cos^2{x}(\cos^2{x}-1)+\cos^2{x}-1)}{x^2(1+\cos{x}\sqrt{\cos{2x}})}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{-(2\cos^2{x}(\cos^2{x}-1)+1(\cos^2{x}-1))}{x^2(1+\cos{x}\sqrt{\cos{2x}})}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{-(\cos^2{x}-1)(2\cos^2{x}+1)}{x^2(1+\cos{x}\sqrt{\cos{2x}})}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{(1-\cos^2{x})(2\cos^2{x}+1)}{x^2(1+\cos{x}\sqrt{\cos{2x}})}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin^2{x}(2\cos^2{x}+1)}{x^2(1+\cos{x}\sqrt{\cos{2x}})}}$

Factorize the Algebraic trigonometric function

Now, factorise this algebraic trigonometric function as two functions in the following way.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[\dfrac{\sin^2{x}}{x^2} \times \dfrac{2\cos^2{x}+1}{1+\cos{x}\sqrt{\cos{2x}}}\Bigg]}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[{\Bigg(\dfrac{\sin{x}}{x}\Bigg)}^2 \times \dfrac{2\cos^2{x}+1}{1+\cos{x}\sqrt{\cos{2x}}}\Bigg]}$

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

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

Evaluate the Limit of the each function

According to the limit rule of trigonometric function, the limit of the $\dfrac{\sin{x}}{x}$ as $x$ tends to $0$ is equal to $1$.

$= \,\,\,$ ${(1)}^2$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2\cos^2{x}+1}{1+\cos{x}\sqrt{\cos{2x}}}}$

$= \,\,\,$ $1 \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2\cos^2{x}+1}{1+\cos{x}\sqrt{\cos{2x}}}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2\cos^2{x}+1}{1+\cos{x}\sqrt{\cos{2x}}}}$

Now, find the limit of the trigonometric function as x approaches 0 by the direct substitution method.

$= \,\,\,$ $\dfrac{2\cos^2{(0)}+1}{1+\cos{(0)}\sqrt{\cos{2(0)}}}$

The value of cos of zero is equal to one.

$= \,\,\,$ $\dfrac{2{(1)}^2+1}{1+1\sqrt{\cos{(0)}}}$

$= \,\,\,$ $\dfrac{2 \times 1+1}{1+\sqrt{1}}$

$= \,\,\,$ $\dfrac{2+1}{1+1}$

$= \,\,\,$ $\dfrac{3}{2}$

$\therefore \,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1-\cos{x}\sqrt{\cos{2x}}}{x^2}}$ $\,=\,$ $\dfrac{3}{2}$

Thus, the limit of the algebraic trigonometric function is evaluated in two different methods in calculus.

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