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Tyler's question at Yahoo! Answers involving the limit of the arctangent of an indeterminate form

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MarkFL

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Feb 24, 2012
13,775
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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:

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