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$\lim_{x\to-3.01}\frac{\tan\left(x+3\right)}{x+3}$

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Other solved exercises

$\lim_{x\to-\infty}\left(2^x+x^2+x\right)$

$-28^0$

$-\frac{1}{5}+\left(-\infty\right)$

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$-10+3x<10$

$\left(13ab^3-5b\:\right)^2$

$16a^2-40a+25$
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