Math Doubts

Evaluate $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{(1-\cos{2x})(3+\cos{x})}{x\tan{4x}}}$

The limit of the quotient of the product of $1-\cos{2x}$ and $3+\cos{x}$ by $x\tan{4x}$ has to evaluate as $x$ approaches $0$ in this limit problem, where $x$ represents an angle of right triangle and also a variable.

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

Firstly, let us evaluate the given algebraic trigonometric function direct substitution method.

$= \,\,\,$ $\dfrac{(1-\cos{2(0)})(3+\cos{(0)})}{0 \times \tan{4(0)}}$

$= \,\,\,$ $\dfrac{(1-\cos{(0)})(3+\cos{(0)})}{0 \times \tan{(0)}}$

According to the trigonometry, the value of cosine of angle zero is equal to $1$ and the value of tan of angle zero is equal to $0$.

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

$= \,\,\,$ $\dfrac{(0)(4)}{0}$

$= \,\,\,$ $\dfrac{0 \times 4}{0}$

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

The direct substitution method is failed to calculate the limit of the given function as $x$ approaches $0$. So, this limit problem has to do in another way.

Simplify the Algebraic trigonometric function

According to the power reduction trigonometric identity,

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

So, the trigonometric expression $1-\cos{2x}$ can be replaced by its equivalent value $2\sin^2{x}$ in the given function.

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

Now, split the function as the following two functions for simplifying the given function further.

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

According to the product rule of limits, the limit of product of two functions $\dfrac{2\sin^2{x}}{x\tan{4x}}$ and $3+\cos{x}$ can be evaluated by the product of their limits.

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

Evaluate the Limit of trigonometric function

Look at the second limit of the function and its value can be evaluated by direct substitution method but do not disturb the first limit function.

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

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

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2\sin^2{x}}{x\tan{4x}}}$ $\times$ $4$

$= \,\,\,$ $4$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2\sin^2{x}}{x\tan{4x}}}$

Simplify the Limit of the function further

In the algebraic trigonometric function, $2$ is a constant and it can be separated from the function by the constant multiple limit rule.

$= \,\,\,$ $4 \times 2$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin^2{x}}{x\tan{4x}}}$

$= \,\,\,$ $8 \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin^2{x}}{x\tan{4x}}}$

Now, split the function as the product of two functions due to the involvement of sine and tan functions.

$= \,\,\,$ $8 \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{\sin^2{x}}{x} \times \dfrac{1}{\tan{4x}}\Bigg)}$

Each factor contains a trigonometric function. So, the trigonometric limit rules have to use to find the limit of the product of the functions. Therefore, let us try to adjust each factor same as the limit rules of trigonometric functions sin and tan functions.

$= \,\,\,$ $8 \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{\sin^2{x}}{x} \times \dfrac{1}{\tan{4x}} \times 1\Bigg)}$

$= \,\,\,$ $8 \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{\sin^2{x}}{x} \times \dfrac{1}{\tan{4x}} \times \dfrac{x}{x}\Bigg)}$

$= \,\,\,$ $8 \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{\sin^2{x}}{x} \times \dfrac{1}{\tan{4x}} \times x \times \dfrac{1}{x}\Bigg)}$

$= \,\,\,$ $8 \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{\sin^2{x}}{x} \times \dfrac{1}{x} \times \dfrac{1}{\tan{4x}} \times x\Bigg)}$

$= \,\,\,$ $8 \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{\sin^2{x} \times 1}{x \times x} \times \dfrac{1 \times x}{\tan{4x}}\Bigg)}$

$= \,\,\,$ $8 \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{\sin^2{x}}{x^2} \times \dfrac{1}{\dfrac{\tan{4x}}{x}}\Bigg)}$

$= \,\,\,$ $8 \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin^2{x}}{x^2}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\tan{4x}}{x}}\Bigg)}$

$= \,\,\,$ $8 \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin^2{x}}{x^2}}$ $\times$ $\dfrac{1}{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\tan{4x}}{x}}}\Bigg)$

$= \,\,\,$ $8 \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin^2{x}}{x^2}}$ $\times$ $\dfrac{1}{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize 1 \times \dfrac{\tan{4x}}{x}}}\Bigg)$

$= \,\,\,$ $8 \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin^2{x}}{x^2}}$ $\times$ $\dfrac{1}{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{4}{4} \times \dfrac{\tan{4x}}{x}}}\Bigg)$

$= \,\,\,$ $8 \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin^2{x}}{x^2}}$ $\times$ $\dfrac{1}{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{4}{4} \times \dfrac{\tan{4x}}{x}}}\Bigg)$

$= \,\,\,$ $8 \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin^2{x}}{x^2}}$ $\times$ $\dfrac{1}{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize 4 \times \dfrac{1}{4} \times \dfrac{\tan{4x}}{x}}}\Bigg)$

$= \,\,\,$ $8 \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin^2{x}}{x^2}}$ $\times$ $\dfrac{1}{4 \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{4} \times \dfrac{\tan{4x}}{x}}}\Bigg)$

$= \,\,\,$ $8 \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin^2{x}}{x^2}}$ $\times$ $\dfrac{1}{4 \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1 \times \tan{4x}}{4 \times x}}}\Bigg)$

$= \,\,\,$ $8 \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin^2{x}}{x^2}}$ $\times$ $\dfrac{1}{4 \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\tan{4x}}{4x}}}\Bigg)$

$= \,\,\,$ $8 \times \dfrac{1}{4} \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin^2{x}}{x^2}}$ $\times$ $\dfrac{1}{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\tan{4x}}{4x}}}\Bigg)$

$= \,\,\,$ $\dfrac{8}{4} \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin^2{x}}{x^2}}$ $\times$ $\dfrac{1}{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\tan{4x}}{4x}}}\Bigg)$

$= \,\,\,$ $\require{cancel} \dfrac{\cancel{8}}{\cancel{4}} \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin^2{x}}{x^2}}$ $\times$ $\dfrac{1}{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\tan{4x}}{4x}}}\Bigg)$

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

Evaluate the Limits of the functions

Now, let’s try to evaluate the limit of each function by using trigonometric limit rules.

The powers of expressions in numerator and denominator are same. So, it can be written as follows by the power of a quotient rule.

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

According to limit rule of an exponential function, it can be simplified further.

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

According to a standard result, the limit of sinx/x as x approaches 0 is equal to $1$.

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

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

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

If $x \to 0$, then $4 \times x \to 4 \times 0$. Therefore, $4x \to 0$. It is cleared that if $x$ approaches zero, then $4x$ also approaches zero.

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

Take $y = 4x$, then express the function in terms of $y$.

$= \,\,\,$ $2 \times \Bigg(\dfrac{1}{\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\tan{y}}{y}}}\Bigg)$

According to another standard result, the limit of tanx/x as x approaches 0 is equal to $1$. Therefore, the limit of $\dfrac{\tan{y}}{y}$ as $y$ approaches $0$ is also equal to $1$.

$= \,\,\,$ $2 \times \Bigg(\dfrac{1}{1}\Bigg)$

$= \,\,\,$ $2 \times 1$

$= \,\,\, 2$

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