If $f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$,then $\lim_{\alpha \rightarrow 0} \frac{f(1-\alpha) - f(1)}{\alpha^3 + 3\alpha} = $

$\lim _{x \rightarrow 0} \left( \frac{x \cdot 10^x - x}{1 - \cos x} \right)$ is equal to

$\lim _{x \rightarrow 0} \frac{\cos 2x - \cos 3x}{\cos 4x - \cos 5x} = $

Let $f(x) = \lim_{n \rightarrow \infty} \sum_{r=0}^n \left( \frac{2\tan(x/2^{r+1})}{1 - \tan^2(x/2^{r+1})} \right)$. Then $\lim_{x \rightarrow 0} \frac{e^x - e^{f(x)}}{x - f(x)}$ is equal to . . . . . . .

Evaluate the limit: $\lim _{x \rightarrow 0^{+}}\left(e^{x}+x\right)^{1 / x}$

If $f(x) = 3x^{10} - 7x^8 + 5x^6 - 21x^3 + 3x^2 - 7$,then $\mathop {\lim }\limits_{\alpha \to 0} \frac{f(1 - \alpha) - f(1)}{\alpha^3 + 3\alpha}$ is