MHB Limit of Arctan (x^2-4)/(3x^2-6x) as x Approaches 2

[MarkFL]

Here is the question:

What is the limit of arctan( (x^2-4) / (3x^2-6x) ) as x approaches 2?

What is the limit of arctan( (x^2-4) / (3x^2-6x) ) as x approaches 2

Here is a link to the question:

What is the limit of arctan( (x^2-4) / (3x^2-6x) ) as x approaches 2? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello Tyler,

we are given to evaluate:

$\displaystyle \lim_{x\to 2}\tan^{-1}\left(\frac{x^2-4}{3x^2-6x} \right)$

We observe that substituting 2 for $x$ gives us the indeterminate form 0/0 for the function's argument.

If we factor the numerator and denominator of the argument for the inverse tangent function, we obtain:

$\displaystyle \lim_{x\to 2}\tan^{-1}\left(\frac{(x+2)(x-2)}{3x(x-2)} \right)$

Dividing out common factors:

$\displaystyle \lim_{x\to 2}\tan^{-1}\left(\frac{x+2}{3x} \right)$

Now, we no longer have an indeterminate form and we may use substitution to obtain the value of the limit:

$\displaystyle \tan^{-1}\left(\frac{2+2}{3(2)} \right)=\tan^{-1}\left(\frac{2}{3} \right)$