$f\left(x\right)=\frac{x+1}{1\frac{1}{x+1}}$
$\lim_{x\to0}\left(\frac{e^x-\sin\left(x\right)-1}{1-\cos\left(x\right)}\right)$
$\infty^{\frac{1}{2}}$
$-169m^4+49n^2$
$81-18x^2+x^4$
$\lim_{x\to\infty}\sec\left(x\right)$
$\lim_{x\to-2}\left(\frac{x+3}{x+2}\right)$
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