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

$x$ is angle of right triangle in radian and it formed an algebraic trigonometric expression as the limit $x$ approaches zero.

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

Substitute $x = 0$ to study its functionality.

$= \,\,\,$ $\dfrac{\cos^2{(0)}-\sin^2{(0)}-1}{\sqrt{{(0)}^2+1}-1}$

$= \,\,\,$ $\dfrac{{(1)}^2-{(0)}^2-1}{\sqrt{0+1}-1}$

$= \,\,\,$ $\dfrac{1-0-1}{\sqrt{1}-1}$

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

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

The value of algebraic trigonometric function is undefined. So, its value cannot be evaluated by the direct substitution method. So, it should be solved in another mathematical approach.

Simplify trigonometric expression in numerator

The numerator is a trigonometric expression and the denominator is an algebraic expression. So, try to simplify the expression in the numerator firstly. The trigonometric expression contains three terms in which two terms are trigonometric functions. So, transform one trigonometric function as another trigonometric function but it is better to transform cos function as sin function. The only reason is, there is a limit formula in sin function.

The square of cos function can be written in terms of square of sin function by the Pythagorean identity of sin and cos functions.

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

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

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0} \normalsize \require{cancel} \dfrac{\cancel{1}-2\sin^2{x}-\cancel{1}}{\sqrt{x^2+1}-1}$

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

The trigonometric expression in the numerator is successfully simplified and it is time to simplify the algebraic expression in the denominator.

Simplify algebraic expression in denominator

The algebraic expression can be simplified by using rationalizing method.

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

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

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

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

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

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

Step to evaluate the function as limit x approaches 0

The entire algebraic trigonometric function is simplified successfully and it is time to find the value of the function as the value of $x$ approaches zero.

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

The sin function is in square form and the algebraic function is also in square form. Separate the function as two multiplying factors.

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

Use product rule of limit for applying limit to both multiplying factors.

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

Find value of algebraic function as limit x tends to 0

Substitute $x = 0$ in the second multiplying function to find the value of algebraic expression as the limit $x$ approaches zero.

$= \,\,\,$ $-2 \Bigg(\displaystyle \large \lim_{x \,\to\, 0} \normalsize {\Bigg[\dfrac{\sin{x}}{x}\Bigg]}^2$ $\times$ $(\sqrt{{(0)}^2+1}+1) \Bigg)$

$= \,\,\,$ $-2 \Bigg(\displaystyle \large \lim_{x \,\to\, 0} \normalsize {\Bigg[\dfrac{\sin{x}}{x}\Bigg]}^2$ $\times$ $(\sqrt{0+1}+1)\Bigg)$

$= \,\,\,$ $-2 \Bigg(\displaystyle \large \lim_{x \,\to\, 0} \normalsize {\Bigg[\dfrac{\sin{x}}{x}\Bigg]}^2$ $\times$ $(\sqrt{1}+1)\Bigg)$

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

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

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

Find value of trigonometric function as limit x tends to 0

The expression is almost similar to the ratio of sin of angle to angle as limit angle approaches zero rule. So, try to adjust the expression to apply the limit formula.

$= \,\,\,$ $-4 \displaystyle \large \lim_{x \,\to\, 0} \normalsize {\Bigg[\dfrac{\sin{x}}{x}\Bigg]}^2$

$= \,\,\,$ $-4 {\Bigg[\displaystyle \large \lim_{x \,\to\, 0} \normalsize \dfrac{\sin{x}}{x}\Bigg]}^2$

As per limit of ratio of sinx to x as x approaches 0 rule, the value of the function is one.

$= \,\,\,$ $-4 \times {(1)}^2$

$= \,\,\,$ $-4 \times 1$

$= \,\,\,$ $-4$