Given,

$$\begin{aligned} \lim _{x \rightarrow 0} \frac{\sin x-2 \sin 3 x+\sin 5 x}{x} & =\lim _{x \rightarrow 0} \frac{\sin 5 x+\sin x-2 \sin 3 x}{x} \\ & =\lim _{x \rightarrow 0} \frac{2 \sin 3 x \cos 2 x-2 \sin 3 x}{x}=\lim _{x \rightarrow 0} \frac{2 \sin 3 x(\cos 2 x-1)}{x} \\ & =\lim _{x \rightarrow 0} \frac{2 \sin 3 x}{\frac{1}{3} \times 3 x}(\cos 2 x-1)=6 \lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x}(\cos 2 x-1) \\ & =6 \lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x} \cdot \lim _{x \rightarrow 0}(\cos 2 x-1)=6 \times 1 \times 0=0 \end{aligned}$$

$$\begin{aligned} & \text { Given, } \quad \lim _{x \rightarrow 1} \frac{x^4-1}{x-1}=\lim _{x \rightarrow k} \frac{x^3-k^3}{x^2-k^2} \\ & \Rightarrow \quad 4(1)^{4-1}=\lim _{x \rightarrow k} \frac{\frac{x^3-k^3}{x-k}}{\frac{x^2-k^2}{x-k}} \quad\left[\begin{array}{l} \because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a} \\ =n a^{n-1} \end{array}\right] \end{aligned}$$

$$\begin{array}{ll} \Rightarrow & 4=\frac{\lim _\limits{x \rightarrow k} \frac{x^3-k^3}{x-k}}{\lim _\limits{x \rightarrow k} \frac{x^2-k^2}{x-k}} 1 \quad\left[\because \lim _\limits{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{\lim _\limits{x \rightarrow a} f(x)}{\lim _\limits{x \rightarrow a} g(x)}\right]\\ \Rightarrow & 4=\frac{3 k^2}{2 k} \Rightarrow 4=\frac{3}{2} k \\ \therefore & k=\frac{4 \times 2}{3}=\frac{8}{3} \end{array}$$

$$\begin{aligned} & \frac{d}{d x}\left(\frac{x^4+x^3+x^2+1}{x}\right)=\frac{d}{d x}\left(x^3+x^2+x+\frac{1}{x}\right) \\ & =\frac{d}{d x} x^3+\frac{d}{d x} x^2+\frac{d}{d x} x+\frac{d}{d x}\left(\frac{1}{x}\right) \\ & =3 x^2+2 x+1+\left(-\frac{1}{x^2}\right) \\ & =3 x^2+2 x+1-\frac{1}{x^2} \\ & =\frac{3 x^4+2 x^3+x^2-1}{x^2} \end{aligned} $$

$$\begin{aligned} \text { Let } \quad y & =\left(x+\frac{1}{x}\right)^3 \\ \therefore \quad \frac{d y}{d x}=\frac{d}{d x}\left(x+\frac{1}{x}\right)^3 & =3\left(x+\frac{1}{x}\right)^{3-1} \frac{d}{d x}\left(x+\frac{1}{x}\right) \quad \text{[by chain rule]}\\ & =3\left(x+\frac{1}{x}\right)^2\left(1-\frac{1}{x^2}\right) \\ & =3 x^2-\frac{3}{x^2}-\frac{3}{x^4}+3 \end{aligned}$$

$$\begin{aligned} &\text { Let } \quad y=(3 x+5)(1+\tan x)\\ &\therefore \quad \frac{d y}{d x}=\frac{d}{d x}[(3 x+5)(1+\tan x)] \end{aligned}$$

$$\begin{aligned} &\begin{aligned} & =(3 x+5) \frac{d}{d x}(1+\tan x)+(1+\tan x) \frac{d}{d x}(3 x+5) \quad \text { [by product rule]}\\ & =(3 x+5)\left(\sec ^2 x\right)+(1+\tan x) \cdot 3 \\ & =(3 x+5) \sec ^2 x+3(1+\tan x) \\ & =3 x \sec ^2 x+5 \sec ^2 x+3 \tan x+3 \end{aligned}\\ \end{aligned}$$