Math Doubts

Evaluate $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x^3\sin{x}}{(\sec{x}-\cos{x})^2}}$

The algebraic function $x^3$ and trigonometric functions $\sin{x}$, $\cos{x}$ and $\sec{x}$ are formed a rational expression. The limit of the rational expression that contains both algebraic and trigonometric functions, has to evaluate as $x$ approaches $0$ in this limit problem.

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

Simplify the Rational expression

There is only one term in expression of the numerator but there are two terms in the expression of the denominator. It means that we cannot simplify the numerator further but we can simplify the denominator by the trigonometric identities.

In denominator, there are two terms inside the square and they both are reciprocals. So, one trigonometric function can be expressed in its multiplicative inverse form. This mathematical technique helps us in simplify the whole rational expression in the next few steps. Therefore, write the secant function in its reciprocal form by the reciprocal identity of cos function.

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

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

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

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

The subtraction cos squared function from one can be simplified by the sine squared formula.

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

The expression in the denominator can be simplified further by the power of a quotient rule.

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

The square of sin squared function can be simplified by the power rule of exponents.

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

Split the expression as two functions

It is required to divide the rational expression as the product of two fractions.

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

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

Now, use the reciprocal trigonometric identity of cos function for writing its reciprocal as secant squared function.

$=\,\,\,$ $\large \displaystyle \lim_{x \,\to\, 0}{\normalsize \dfrac{x^3\sin{x}}{\sin^4{x} \times \sec^2{x}}}$

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

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

Now, use the multiplicative inverse trigonometric identity to transform the reciprocal of secant function as cosine.

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

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

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

The limiting operation of the whole function can be split to its fractions by the product rule of limits.

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

Evaluate the Limits of the functions

Now, we can evaluate the limit of the second function by the direct substitution method but keep the first factor as it is in this expression.

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

According to trigonometry, the cosine of $0$ radian is equal to $1$.

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

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

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

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

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

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

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

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

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

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

As per the limit rules of trigonometric functions, the limit of sinx by x as x approaches 0 is equal to 1.

$=\,\,\,$ $\Bigg(\dfrac{1}{1}\Bigg)^3$

$=\,\,\,$ $(1)^3$

$=\,\,\,$ $1^3$

$=\,\,\, 1$

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