$\lim _{x \rightarrow \infty} \frac{e^{x^4}-1}{e^{x^4}+1} = $

$\lim _{x \rightarrow 0} \frac{2x}{|x|+x^2} = $

$\lim _{x \rightarrow 0} \frac{\sqrt{1-\cos x^2}}{1-\cos x} = $

$\mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} + 8x + 3} - \sqrt {{x^2} + 4x + 3} ) = $

દ્વિઘાત સમીકરણ જેના બીજ $l$ અને $m$ છે,જ્યાં
$\begin{aligned}
& l=\lim _{\theta \rightarrow 0}\left(\frac{3 \sin \theta-4 \sin ^2 \theta}{\theta}\right), \\
& m=\lim _{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta\left(1-\tan ^2 \theta\right)}, \text{ તે છે}
\end{aligned}$

$\lim _{n \rightarrow \infty}\left(1+\frac{1+\frac{1}{2}+\ldots+\frac{1}{n}}{n^{2}}\right)^{n} = \dots$