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Evaluate $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin^3{x}}{\sin{x}-\tan{x}}}$

A rational expression is formed by the trigonometric functions sine and tangent. In this trigonometric limit problem, the limit of the trigonometric rational expression has to evaluate as $x$ approaches $0$.

$\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin^3{x}}{\sin{x}-\tan{x}}}$

Eliminate common factors from Rational expression

The expression in the numerator is in sine function. There is a sine function in each term of the denominator. So, let’s try to cancel the common sine function in the rational expression.

It can be started by expressing the tan function in rational form of the trigonometric functions.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin^3{x}}{\sin{x}-\dfrac{\sin{x}}{\cos{x}}}}$

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

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

The sine function can be taken common from the terms in the denominator.

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

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

The sine cubed function can be split as two factors for canceling the common factors in the rational function.

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

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

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

Simplify the expression in Rational form

The difference of the terms in the denominator can be simplified by the subtraction of the fractions.

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

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

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

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

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

The square of sine function can be expanded in terms of cosine by the sine squared identity.

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

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

The difference of the squares can be factored and it helps us to cancel the common factors in the rational expression.

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

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

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

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\big(1+\cos{x}\big)\cancel{\big(1-\cos{x}\big)} \times \cos{x}}{-1 \times \cancel{\big(1-\cos{x}\big)}}}$

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

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

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize -\big(1+\cos{x}\big)\cos{x}}$

Evaluate the expression by direct substitution

The trigonometric expression can be evaluated as $x$ closer to $0$ by the direct substitution method.

$=\,\,\,$ $-\big(1+\cos{(0)}\big)\cos{(0)}$

According to the trigonometry, the cosine of angle zero radian is equal to one. Now, simplify the expression for evaluating the limit of the given trigonometric rational expression as $x$ approaches to $0$.

$=\,\,\,$ $-\big(1+1\big) \times 1$

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

$=\,\,\,$ $-2$

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