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

AI Thread Summary
The limit of arctan((x^2-4)/(3x^2-6x)) as x approaches 2 results in an indeterminate form of 0/0. By factoring the numerator and denominator, the expression simplifies to arctan((x+2)/(3x) after canceling common factors. Substituting x = 2 into this simplified expression yields arctan(2/3). Therefore, the limit is arctan(2/3). This calculation confirms the limit's value as x approaches 2.
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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:

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