$\lim _{x \rightarrow 0} \frac{1-\cos (1-\cos x)}{\sin ^4 x} = $

$\lim _{n}$ ${\rightarrow \infty} \frac{\left(1^2-1\right)(n-1)+\left(2^2-2\right)(n-2)+\ldots +\left((n-1)^2-(n-1)\right) \cdot 1}{\left(1^3+2^3+\ldots +n^3\right)-\left(1^2+2^2+\ldots +n^2\right)}$ का मान ज्ञात कीजिए:

यदि $f(x) = \frac{2}{x - 3}$,$g(x) = \frac{x - 3}{x + 4}$ और $h(x) = - \frac{2(2x + 1)}{x^2 + x - 12}$ है,तो $\lim_{x \to 3} [f(x) + g(x) + h(x)]$ का मान ज्ञात कीजिए।

$\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n k^2 x = $

$\mathop {\lim }\limits_{x \to \infty } \frac{{2{x^2} - 3x + 1}}{{{x^2} - 1}} = $

$\lim _{x \rightarrow 0} \frac{e^{x^2}-\cos 3 x}{\sin x \log (1+2 x)}=$