Math Doubts

Evaluate $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\log_{\displaystyle e}{\big(\cos{(\sin{x})}\big)}}{x^2}}$

In this limit problem, the composition of a logarithmic function and two trigonometric functions formed a rational expression with an algebraic function in terms of $x$.

$\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\log_{\displaystyle e}{\big(\cos{(\sin{x})}\big)}}{x^2}}$

The limit of the quotient of natural logarithm of cosine of sine of angle $x$ by $x$ squared should be calculated as $x$ tends to zero and it can evaluated in two different methods. Now, let us learn each method to find the limit of the given rational function.

Fundamental method

Firstly, let’s get started to learn how to find the limit of the rational function by using the limit rules.

Find the limit by the direct substitution

First of all, let us try to find the limit of the given function by the direct substitution method.

$=\,\,\,$ $\dfrac{\log_{\displaystyle e}{\Big(\cos{\big(\sin{(0)}\big)}\Big)}}{(0)^2}$

The sine of angle zero radian is equal to zero as per the trigonometry.

$=\,\,\,$ $\dfrac{\log_{\displaystyle e}{\big(\cos{(0)}\big)}}{0}$

According to trigonometry, the cosine of angle zero radian is equal to one.

$=\,\,\,$ $\dfrac{\log_{\displaystyle e}{\big(1\big)}}{0}$

The natural logarithm of one is zero as per the logarithms.

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

The values of both numerator and denominator are zero. It means, the limit of the given function is indeterminate as the value of $x$ is closer zero. It indicates that we have to think for an alternative method for calculating the limit.

Adjust the log function to required form

A logarithmic function is involved in the numerator of the given rational function. In calculus, there is a limit rule in terms of log function. Hence, it is recommendable to adjust the given function in the form of the log limit rule. It can be initiated by adding $1$ to the trigonometric function and subtracting $1$ from the trigonometric expression, inside the logarithmic function.

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

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

For applying the natural logarithmic limit rule to the logarithmic function, the expression inside the logarithm should be in the denominator. So, multiply the rational function with $\cos{(\sin{x})}-1$ and also divide the same function by the same expression.

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

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

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

The limit of the product can be calculated by calculating the product of their limits as per the product rule of the limits.

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

Evaluate the Limit by the product of the limits

The first factor in the expression can be calculated by using the logarithmic limit rule but it is not in the form of that rule. Hence, let’s try to transform the first factor same as the log limit rule by some acceptable adjustments.

  1. If $x\,\to\,0$ then $\sin{x}\,\to\,\sin{(0)}$. Therefore, $\sin{x}\,\to\,0$
  2. If $\sin{x}\,\to\,0$ then $\cos{(\sin{x})}\,\to\,\cos{(0)}$. Therefore, $\cos{(\sin{x})}\,\to\,1$
  3. If $\cos{(\sin{x})}\,\to\,1$ then $\cos{(\sin{x})}-1\,\to\,1-1$. Therefore, $\cos{(\sin{x})}-1\,\to\,0$

The three steps have cleared that the value of the expression $\cos{(\sin{x})}-1$ is closer to $0$ when the value of $x$ approaches to $0$. Hence, the input of the limit function in first factor can be adjusted as follows but keep the second factor as it is.

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

For avoiding the confusion, let’s take $y \,=\, \cos{(\sin{x})}-1$

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

According to the logarithmic limit rule, the limit of the logarithm of one plus $y$ by $y$ as $y$ approaches $0$ is equal to one.

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

It is time to concentrate on finding the limit of the remaining function.

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

The trigonometric expression in the numerator represents one minus cosine of angle trigonometric formula and it is useful to convert the expression into sine function. There is a limit rule in terms of sine. Hence, it is recommendable to use the combination of both formulas for evaluating the limit.

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

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

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

Now, separate the constant factor from the limit operation as per the constant multiple limit rule.

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

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

The trigonometric expression in the numerator can be converted by using one minus sine of angle trigonometric identity.

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

The rational expression in terms of sine function is not actually same as the trigonometric limit rule in sine function. So, make some acceptable adjustments to prepare the function same as that trigonometric limit rule. In order to get it, multiply the trigonometric function by the function that represents an angle in the sine function. Similarly, divide the function by the same function to maintain the mathematical balance.

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

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

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

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

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

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

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

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

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

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

$=\,\,\,$ $-\displaystyle \large \lim_{x\,\to\,0}{\normalsize \bigg(\dfrac{\cancel{2}}{\cancel{4}} \times \dfrac{\sin^2{\Big(\dfrac{\sin{x}}{2}\Big)}}{x^2}}$ $\times$ $\dfrac{\sin^2{x}}{\Big(\dfrac{\sin{x}}{2}\Big)^2}\bigg)$

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

Once again, use the constant multiple limit rule to release the constant from the limit operation.

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

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

Now, use the multiplication of fractions process to chance their positions in the denominators.

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

Use the product rule of the limits one more time to find the limit of product of two functions by the product of their limits.

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

For our convenience, take $z \,=\, \dfrac{\sin{x}}{2}$

$(1).\,\,\,$ If $x\,\to\,0$ then $\sin{x}\,\to\,\sin{(0)}$. Therefore, $\sin{x}\,\to\,0$

$(2).\,\,\,$ If $\sin{x}\,\to\,0$ then $\dfrac{\sin{x}}{2}\,\to\,\dfrac{0}{2}$. Therefore, $\dfrac{\sin{x}}{2}\,\to\,0$

$(3).\,\,\,$ If $\dfrac{\sin{x}}{2}\,\to\,0$ then $z\,\to\,0$.

The three steps process have cleared that the value of $z$ approaches $0$ when the value of $x$ tends to $0$. Hence, convert the first trigonometric function in terms of $z$ and no need to disturb the second trigonometric function.

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

Each rational function can be simplified by the power of a quotient rule.

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

Now, use the power rule of the limits to find the limit of an exponential form function.

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

According to the trigonometric limit rule in sine function, the limit each rational function is equal to one as the value of its input approaches to zero.

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

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

$=\,\,\,$ $-\dfrac{1}{2}(1)$

$=\,\,\,$ $-\dfrac{1}{2} \times 1$

$=\,\,\,$ $-\dfrac{1}{2}$

L’Hopital’s Rule

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

According to the direct substitution, it is proved that the limit of the given rational function is indeterminate as the value of $x$ approaches to $0$. Hence, the l’hospital’s rule can be used to find the limit by the differentiation.

Differentiate both math expressions

Now, differentiate the logarithmic function with respect to $x$ in the numerator and also differentiate the algebraic function with respect to $x$ in denominator.

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

The expression in the numerator is formed by the composition of the logarithmic and trigonometric functions. Hence, the differentiation of composite function should be done by the chain rule. In this process, the derivative rule of log function, derivative rule of cosine and derivative rule of sine are used in differentiating the expression in the numerator. Similarly, the derivative of the algebraic function is evaluated by the power rule of the differentiation.

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

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

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

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

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

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

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

According to the quotient rule of sine and cosine, the quotient of sine by cosine is tan of the angle

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

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

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

Evaluate the Limit by Direct substitution

Now, substitute $x \,=\, 0$ in both numerator and denominator to find the limit of the rational function as the value of $x$ is closer to zero.

$=\,\,\,$ $\dfrac{(-1)}{2} \times \dfrac{\tan{\big(\sin{(0)}\big)} \times \cos{(0)}}{0}$

According to the trigonometry, sine of angle zero radian is zero and the cosine of angle zero radian is equal to one. Substitute them in the numerator.

$=\,\,\,$ $\dfrac{(-1)}{2} \times \dfrac{\tan{(0)} \times 1}{0}$

The tan of angle zero radian is zero and substitute it in the numerator to calculate the rational function.

$=\,\,\,$ $\dfrac{(-1)}{2} \times \dfrac{0 \times 1}{0}$

$=\,\,\,$ $\dfrac{(-1)}{2} \times \dfrac{0}{0}$

$=\,\,\,$ $\dfrac{(-1) \times 0}{2 \times 0}$

Use the multiplication of fractions system for multiplying the fractions and then find their product.

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

It is evaluated that the limit of the rational function is indeterminate as the value of $x$ tends to zero. So, it is not recommendable to use the direct substitution method at this stage.

Use L’Hospital’s rule one more time

The limit of the given rational function is indeterminate even though L’Hopital’s rule was used once. It clears that the L’Hopital’s rule should be used one more time. So, differentiate both numerator and denominator one more time with respect to $x$.

$=\,\,\,$ $\dfrac{(-1)}{2} \times \displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\dfrac{d}{dx}{\Big(\tan{(\sin{x})} \times \cos{x}\Big)}}{\dfrac{d}{dx}{(x)}}}$

Two trigonometric functions are multiplied in the numerator. So, differentiate the product of the functions with respect to $x$ as per the product rule of the differentiation in the numerator. Similarly, differentiate the denominator with respect to $x$ by using the derivative rule of a variable.

$=\,\,\,$ $\dfrac{(-1)}{2} \times \displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\tan{(\sin{x})} \times \dfrac{d}{dx}{(\cos{x})}+\cos{x} \times \dfrac{d}{dx}{\tan{(\sin{x})}}}{\dfrac{dx}{dx}}}$

The derivative of tan of sine of angle x should be calculated by the chain rule due to the composition of two trigonometric functions.

$=\,\,\,$ $\dfrac{(-1)}{2} \times \displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\tan{(\sin{x})} \times (-\sin{x})+\cos{x} \times \sec^2{(\sin{x})} \times \dfrac{d}{dx}{(\sin{x})}}{1}}$

$=\,\,\,$ $\dfrac{(-1)}{2} \times \displaystyle \large \lim_{x\,\to\,0}{\normalsize \Big(\tan{(\sin{x})} \times (-\sin{x})+\cos{x} \times \sec^2{(\sin{x})} \times \dfrac{d}{dx}{(\sin{x})}\Big)}$

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

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

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

Use the direct substitution method to find the limit of the given rational function as the value of $x$ is closer to $0$.

$=\,\,\,$ $\dfrac{(-1)}{2} \times \Big(\tan{\big(\sin{(0)}\big)} \times \big(-\sin{(0)}\big)$ $+$ $\cos^2{(0)} \times \sec^2{\big(\sin{(0)}\big)}\Big)$

As per the trigonometry, the sine of angle zero radian is zero and the cosine of angle zero is radian. Substitute them in the above expression to find its value.

$=\,\,\,$ $\dfrac{(-1)}{2} \times \Big(\tan{(0)} \times (0)$ $+$ $(1)^2 \times \sec^2{(0)}\Big)$

$=\,\,\,$ $\dfrac{(-1)}{2} \times \Big(\tan{(0)} \times (0)$ $+$ $1 \times \sec^2{(0)}\Big)$

$=\,\,\,$ $\dfrac{(-1)}{2} \times \Big(\tan{(0)} \times 0$ $+$ $\sec^2{(0)}\Big)$

$=\,\,\,$ $\dfrac{(-1)}{2} \times \Big(0$ $+$ $\sec^2{(0)}\Big)$

$=\,\,\,$ $\dfrac{(-1)}{2} \times \Big(\sec^2{(0)}\Big)$

The tan of angle zero radian is zero and the secant of angle zero radian is one as per the trigonometric mathematics. So, let’s substitute them in the above expression.

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

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

$=\,\,\,$ $\dfrac{(-1)}{2}$

$=\,\,\,$ $-\dfrac{1}{2}$

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