In the given limit problem, $x$ is a variable in radian and represents an angle of a right triangle. The tan of angle is written as $\tan{x}$, cos of double angle is written as $\cos{2x}$ and tan of double angle is written as $\tan{2x}$. They formed a special trigonometric function and we have to calculate the limit of this function as $x$ approaches zero.
$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x\tan{2x}-2x\tan{x}}{(1-\cos{2x})^2}}$
Let us try to find the limit of the given function as $x$ tends to zero by using direct substitution method.
$= \,\,\,$ $\dfrac{(0)\tan{2(0)}-2(0)\tan{(0)}}{(1-\cos{2(0)})^2}$
$= \,\,\,$ $\dfrac{0 \times \tan{(0)}-2 \times 0 \times \tan{(0)}}{(1-\cos{(0)})^2}$
According to trigonometry, tan of angle zero is zero and cos of angle zero is one.
$= \,\,\,$ $\dfrac{0 \times 0-2 \times 0 \times 0}{(1-1)^2}$
$= \,\,\,$ $\dfrac{0-2 \times 0}{(0)^2}$
$= \,\,\,$ $\dfrac{0-0}{0}$
$= \,\,\,$ $\dfrac{0}{0}$
It is indeterminate and it is impossible to find the limit of the given algebraic trigonometric function by the direct substitution method. So, let us try another way to evaluate it.
In numerator, $\tan{2x}$ is a factor but represents tan of double angle function and it can be expanded by using tan of double angle formula.
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x\Bigg(\dfrac{2\tan{x}}{1-\tan^2{x}}\Bigg)-2x\tan{x}}{(1-\cos{2x})^2}}$
Now, simplify the whole function to move ahead in getting the limit of the given algebraic trigonometric function.
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\dfrac{x \times 2\tan{x}}{1-\tan^2{x}}-2x\tan{x}}{(1-\cos{2x})^2}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\dfrac{2x\tan{x}}{1-\tan^2{x}}-2x\tan{x}}{(1-\cos{2x})^2}}$
The expression in the numerator contains two terms but both terms contains $2x\tan{x}$ commonly. So, they can be taken out common from the terms by factorization.
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2x\tan{x}\Bigg(\dfrac{1}{1-\tan^2{x}}-1\Bigg)}{(1-\cos{2x})^2}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2x\tan{x}\Bigg(\dfrac{1-(1-\tan^2{x})}{1-\tan^2{x}}\Bigg)}{(1-\cos{2x})^2}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2x\tan{x}\Bigg(\dfrac{1-1+\tan^2{x}}{1-\tan^2{x}}\Bigg)}{(1-\cos{2x})^2}}$
$= \,\,\,$ $\require{cancel} \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2x\tan{x}\Bigg(\dfrac{\cancel{1}-\cancel{1}+\tan^2{x}}{1-\tan^2{x}}\Bigg)}{(1-\cos{2x})^2}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2x\tan{x}\Bigg(\dfrac{\tan^2{x}}{1-\tan^2{x}}\Bigg)}{(1-\cos{2x})^2}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\dfrac{2x\tan{x} \times \tan^2{x}}{1-\tan^2{x}}}{(1-\cos{2x})^2}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\dfrac{2x\tan^3{x}}{1-\tan^2{x}}}{(1-\cos{2x})^2}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\dfrac{2x\tan^3{x}}{1-\tan^2{x}}}{\dfrac{(1-\cos{2x})^2}{1}}}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[\dfrac{2x\tan^3{x}}{1-\tan^2{x}} \times \dfrac{1}{(1-\cos{2x})^2}\Bigg]}$
Now, we have the limit of multiplication of two trigonometric functions as an expression. A trigonometric function whose limit is finite can be separated from the multiplication of the two trigonometric functions and it is really helpful to us in simplifying the given function.
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[\dfrac{2 \times x\tan^3{x}}{1-\tan^2{x}} \times \dfrac{1}{(1-\cos{2x})^2}\Bigg]}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[\dfrac{2}{1-\tan^2{x}} \times \dfrac{x\tan^3{x} \times 1}{(1-\cos{2x})^2}\Bigg]}$
$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[\dfrac{2}{1-\tan^2{x}} \times \dfrac{x\tan^3{x}}{(1-\cos{2x})^2}\Bigg]}$
The limit of multiplication of the functions can be split as product of their limits as per the product rule of limits.
$= \,\,\,$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2}{1-\tan^2{x}}\Bigg)}$ $\times$ $\Bigg( \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x\tan^3{x}}{(1-\cos{2x})^2}\Bigg)}$
Now, find the limit of the first factor by the direct substitution but do not disturb the second factor.
$= \,\,\,$ $\Bigg(\dfrac{2}{1-\tan^2{(0)}}\Bigg)$ $\times$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x\tan^3{x}}{(1-\cos{2x})^2}\Bigg)}$
$= \,\,\,$ $\Bigg(\dfrac{2}{1-(0)^2}\Bigg)$ $\times$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x\tan^3{x}}{(1-\cos{2x})^2}\Bigg)}$
$= \,\,\,$ $\Bigg(\dfrac{2}{1-0}\Bigg)$ $\times$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x\tan^3{x}}{(1-\cos{2x})^2}\Bigg)}$
$= \,\,\,$ $\Bigg(\dfrac{2}{1}\Bigg)$ $\times$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x\tan^3{x}}{(1-\cos{2x})^2}\Bigg)}$
$= \,\,\,$ $\Big(2\Big)$ $\times$ $\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x\tan^3{x}}{(1-\cos{2x})^2}\Bigg)}$
$= \,\,\,$ $2 \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x\tan^3{x}}{(1-\cos{2x})^2}\Bigg)}$
$= \,\,\,$ $2\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x\tan^3{x}}{(1-\cos{2x})^2}\Bigg)}$
Look at the expression in the denominator, the expression $1-\cos{2x}$ is the subtraction of cos double angle from one and it can be simplified by a trigonometric identity.
$= \,\,\,$ $2\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x\tan^3{x}}{(2\sin^2{x})^2}\Bigg)}$
It can be simplified further by using the power rule of a product.
$= \,\,\,$ $2\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x\tan^3{x}}{(2)^2 \times (\sin^2{x})^2}\Bigg)}$
Express the value of square of $2$ by exponentiation and also write the value for the square of sin squared of angle $x$ by using the power rule of exponents.
$= \,\,\,$ $2\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x\tan^3{x}}{4 \times \sin^4{x}}\Bigg)}$
$= \,\,\,$ $2\Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{4} \times \dfrac{x\tan^3{x}}{\sin^4{x}}\Bigg)}$
$= \,\,\,$ $2 \times \dfrac{1}{4} \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x\tan^3{x}}{\sin^4{x}}\Bigg)}$
$= \,\,\,$ $\dfrac{2 \times 1}{4} \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x\tan^3{x}}{\sin^4{x}}\Bigg)}$
$= \,\,\,$ $\dfrac{2}{4} \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x\tan^3{x}}{\sin^4{x}}\Bigg)}$
$= \,\,\,$ $\require{cancel} \dfrac{\cancel{2}}{\cancel{4}} \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x\tan^3{x}}{\sin^4{x}}\Bigg)}$
$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x\tan^3{x}}{\sin^4{x}}\Bigg)}$
The algebraic trigonometric function can be simplified by using the basic trigonometric identities.
$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x(\tan{x})^3}{\sin^4{x}}\Bigg)}$
$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x \times \Big(\dfrac{\sin{x}}{\cos{x}}\Big)^3 }{\sin^4{x}}\Bigg)}$
$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x \times \dfrac{\sin^3{x}}{\cos^3{x}} }{\sin^4{x}}\Bigg)}$
$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x \times \dfrac{\sin^3{x} \times 1}{\cos^3{x}} }{\sin^4{x}}\Bigg)}$
$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x \times \sin^3{x} \times \dfrac{1}{\cos^3{x}} }{\sin^4{x}}\Bigg)}$
$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[\dfrac{x \times \sin^3{x} }{\sin^4{x}} \times \dfrac{1}{\cos^3{x}}\Bigg]\Bigg)}$
$= \,\,\,$ $\require{cancel} \dfrac{1}{2} \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[\dfrac{x \times \cancel{\sin^3{x}} }{\cancel{\sin^4{x}}} \times \dfrac{1}{\cos^3{x}}\Bigg]\Bigg)}$
$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[\dfrac{x \times 1 }{\sin{x}} \times \dfrac{1}{\cos^3{x}}\Bigg]\Bigg)}$
$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg[\dfrac{x}{\sin{x}} \times \dfrac{1}{\cos^3{x}}\Bigg]\Bigg)}$
Use the product rule of limits one more time to evaluate the limit of the product of functions by the product of their limits.
$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\Bigg[\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x}{\sin{x}}\Bigg]} \normalsize \times \Bigg[\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\cos^3{x}}\Bigg]\Bigg)}$
$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\Bigg[\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{x}}{x}}\Bigg]} \normalsize \times \Bigg[\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\cos^3{x}}\Bigg]\Bigg)}$
$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\Bigg[\dfrac{1}{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}}{x}}}\Bigg] \normalsize \times \Bigg[\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\cos^3{x}}\Bigg]\Bigg)}$
As per the trigonometric limit rules, the limit of sinx/x as x approaches 0 is equal to one.
$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\Bigg[\dfrac{1}{1}\Bigg] \normalsize \times \Bigg[\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\cos^3{x}}\Bigg]\Bigg)}$
$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\Big[1\Big] \times \Bigg[\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\cos^3{x}}\Bigg]\Bigg)}$
$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(1 \times \Bigg[\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\cos^3{x}}\Bigg]\Bigg)}$
$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\cos^3{x}}\Bigg)}$
Finally, evaluate the limit of the remaining trigonometric function by the direct substitution method.
$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\dfrac{1}{\cos^3{(0)}}\Bigg)$
Actually, the cos of zero radian is equal to one.
$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\dfrac{1}{(1)^3}\Bigg)$
$= \,\,\,$ $\dfrac{1}{2} \times \Bigg(\dfrac{1}{1}\Bigg)$
$= \,\,\,$ $\dfrac{1}{2} \times \Big(1\Big)$
$= \,\,\,$ $\dfrac{1}{2} \times 1$
$= \,\,\,$ $\dfrac{1}{2}$
$= \,\,\,$ $0.5$
Therefore, it is evaluated that the limit of the quotient of subtraction of two times the product of $x$ and $\tan{x}$ from the product of $x$ and $\tan{2x}$ by the square of subtraction of $\cos{2x}$ from $1$ is equal to $\dfrac{1}{2}$ or $0.5$ as $x$ tends to zero.
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