$x$ is an angle of the right angled triangle, $\tan{x}$ is a trigonometric function. A special trigonometric function is formed to represent a value in mathematics as $x$ approaches $\tan^{-1}{3}$.

$\displaystyle \large \lim_{x \,\to\, \tan^{-1}{3}} \normalsize {\dfrac{\tan^2{x}-2\tan{x}-3}{\tan^2{x}-4\tan{x}+3}}$

The function is in terms of $\tan{x}$. So, try to change the limit value in the same form.

If $x \to \tan^{-1}{3}$, then $\tan{x} \to 3$.

$\implies \displaystyle \large \lim_{\tan{x} \,\to\, 3} \normalsize {\dfrac{\tan^2{x}-2\tan{x}-3}{\tan^2{x}-4\tan{x}+3}}$

Substitute $\tan{x} = 3$ to find the value of the function as $\tan{x}$ approaches $3$.

$= \dfrac{{(3)}^2-2 \times 3-3}{{(3)}^2-4 \times 3+3}$

$= \dfrac{9-6-3}{9-12+3}$

$= \dfrac{9-9}{12-12}$

$= \dfrac{0}{0}$

It is undefined. So, the limit of trigonometric function should be solved in another method to obtain value of the function as $x$ approaches $\tan^{-1}{3}$

It is a trigonometric function which contains $\tan{x}$ terms in different forms. The trigonometric function cannot be simplified by only using trigonometric identities. So, alternative method should be searched to evaluate it.

Observe the trigonometric expressions in both numerator and denominator. They both are representing a quadratic equation. So, try to simplify the expression in both numerator and denominator by quadratic equation system.

Express each trigonometric expression as factors by using factoring method of quadratics.

$= \displaystyle \large \lim_{\tan{x} \,\to\, 3} \normalsize {\dfrac{\tan^2{x}-3\tan{x}+\tan{x}-3}{\tan^2{x}-3\tan{x}-\tan{x}+3}}$

$= \displaystyle \large \lim_{\tan{x} \,\to\, 3} \normalsize {\dfrac{\tan{x}{(\tan{x}-3)}+\tan{x}-3}{\tan{x}{(\tan{x}-3)}-\tan{x}+3}}$

$= \displaystyle \large \lim_{\tan{x} \,\to\, 3} \normalsize {\dfrac{\tan{x}{(\tan{x}-3)}+1{(\tan{x}-3)}}{\tan{x}{(\tan{x}-3)}-1{(\tan{x}-3)}}}$

$= \displaystyle \large \lim_{\tan{x} \,\to\, 3} \normalsize {\dfrac{(\tan{x}-3)(\tan{x}+1)}{(\tan{x}-3)(\tan{x}-1)}}$

Now, simplify the function to the possible level.

$= \displaystyle \large \lim_{\tan{x} \,\to\, 3} \normalsize \require{cancel} {\dfrac{\cancel{(\tan{x}-3)}(\tan{x}+1)}{\cancel{(\tan{x}-3)}(\tan{x}-1)}}$

$= \displaystyle \large \lim_{\tan{x} \,\to\, 3} \normalsize {\dfrac{\tan{x}+1}{\tan{x}-1}}$

Substitute $\tan{x} = 3$ to find the value of the function as $\tan{x}$ approaches $3$.

$= \dfrac{3+1}{3-1}$

$= \dfrac{4}{2}$

$= \require{cancel} \dfrac{\cancel{4}}{\cancel{2}}$

$= 2$

It is the solution for this limits problem of trigonometric function.

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