$x$ is a literal but also represents the angle of the right angled triangle in trigonometry. There is a function developed to represent a quantity in mathematical form in terms of $x$ and trigonometric functions sine, cosine and secant.

$\dfrac{x^3\sin{x}}{{(\sec{x}-\cos{x})}^2}$

The value of this algebraic trigonometric function is required to evaluate as the value of $x$ approaches zero. It is written in mathematical form as follows.

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

01

The secant function can be transformed in terms of cosine function on the basis of reciprocal relation between cosine and secant functions. The technique in this math problem is used to transform the portion of trigonometric functions in terms of sine and cosine functions purely.

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

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

The subtraction of square of cosine function from one is equal to square of sine function as per the Pythagorean identity of sine and cosine functions.

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

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

02

Now, simplify the algebraic trigonometric function to obtain the expression in required form for solving this problem.

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

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

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

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

03

The limit value belongs to both multiplying functions. Therefore, it is can be applied to both factors to move further in solving this limit math problem.

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

The first multiplying limit of the function is required to do some adjustments in order to transform the entire function in requisite form but no need to do anything for the second multiplying limit of the function.

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

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

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

According to limit rule for the ratio of sinx to x as x tends to zero, its value is one mathematically.

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

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

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

$=\,\,\, 1$

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