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Use continuity to evaluate the limit.

$$\lim_{x\to 3} \arctan\left(\frac{x^2-9}{5x^2-15x}\right)$$ Factored and got... $$\lim_{x\to 3} \arctan\left(\frac{x+3}{5x}\right)$$ Then... $$\lim_{x\to 3} \arctan\left(\frac{2}{5}\right)$$ Then...my decimal evaluation of $0.5404195$ radians does not solve the problem?

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    $\dfrac{3+3}{5\times3}=$.? – 2017-02-10
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    you made an arithmetic mistake going from 2nd to 3rd expression – 2017-02-10
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    Pick up $\lim$ in last expression. – 2017-02-10
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    That still won't give you an exact, nice-looking decimal expansion for the result; but $\arctan\frac{2}{5}$ **is** a closed-form, valid answer. The decimal expansion of a number is just one way to write it. (If the answer were $1/3$, for instance, would you doubt it because $0.333333\dots$ is not exact?) – 2017-02-10
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    @ClementC. I'm using webassign software to plug in my answers and it's confused because the way I put in the answer so it did get my little confused. – 2017-02-10

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Since $\arctan$ is continuous in $\mathbf R$ we have $$\lim_{x\to 3} \arctan\left(\frac{x+3}{5x}\right)\stackrel{!}{=} \arctan \left(\lim_{x\to 3} \frac{x+3}{5x}\right)= \arctan \left(\frac{2}{5} \right).$$ You just calculated the last limit wrong.