79
If $y=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots$, then $\frac{d y}{d x}=$ .........
Explanation
$$\begin{aligned} \text{Given,}\quad y & =1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots \\ \therefore\quad \frac{d y}{d x} & =0+1+\frac{2 x}{2}+\frac{3 x^2}{6}+\frac{4 x^3}{4!} \\ & =1+x+\frac{x^2}{2}+\frac{x^3}{6}+\ldots \\ & =1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots \\ & =y \end{aligned}$$
80
$\lim _\limits{x \rightarrow 3^{+}} \frac{x}{[x]}=$ .........
Explanation
Given,
$$\begin{aligned} \lim _{x \rightarrow 3^{+}} \frac{x}{[x]} & =\lim _{h \rightarrow 0} \frac{(3+h)}{[3+h]} \\ & =\lim _{h \rightarrow 0} \frac{(3+h)}{3}=1 \end{aligned}$$