$\left(c\right) : We have, a_1, a_2, ...., a_n are in A.P.$

$\therefore a_2- a_1= a_3- a_2=......= a_n- a_{n-1}=d$

$Now, \underset{n\to \infty }{\lim }\sqrt{\frac{d}{n}}(\frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+...+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}})$

$= \underset{n\to \infty }{\lim }\sqrt{\frac{d}{n}}(\frac{\sqrt{a_2}-\sqrt{a_1}}{a_2-a_1}+\frac{\sqrt{a_3}-\sqrt{a_2}}{a_3-a_2}+...+\frac{\sqrt{a_n}-\sqrt{a_{n-1}}}{a_n-a_{n-1}})$

$= \underset{n\to \infty }{\lim }\sqrt{\frac{d}{n}}(\frac{\sqrt{a_2}-\sqrt{a_1}}{d}+\frac{\sqrt{a_3}-\sqrt{a_2}}{d}+...+\frac{\sqrt{a_n}-\sqrt{a_{n-1}}}{d})$

$=\underset{n\to \infty }{\lim }\frac{1}{\sqrt{nd}}(\sqrt{a_n}-\sqrt{a_1})$

$=\underset{n\to \infty }{\lim }\frac{1}{\sqrt{nd}}\left(\frac{a_n-a_1}{\sqrt{a_n}+\sqrt{a_1}}\right)=\underset{n\to \infty }{\lim }\frac{1}{\sqrt{nd}}\frac{(n-1)d}{\sqrt{a_n}+\sqrt{a_1}}$

$=\underset{n\to \infty }{\lim }\frac{1}{\sqrt{nd}}\frac{n(1-{\displaystyle \frac{1}{n}})d}{\sqrt{n}(\sqrt{{\displaystyle \frac{a_1}{n}}+(1-{\displaystyle \frac{1}{n}})}d+\sqrt{{\displaystyle \frac{a_1}{n}}})}$

$=\underset{n\to \infty }{\lim }\frac{(1-{\displaystyle \frac{1}{n}})d}{\sqrt{d}\left(\sqrt{{\displaystyle \frac{a_1}{n}}+d-{\displaystyle \frac{d}{n}}}\right)+\sqrt{{\displaystyle \frac{a_1}{n}}}}=\frac{d}{\sqrt{d}·\sqrt{d}}=1$

(d) : $\left(S1\right) : \underset{n\to \infty }{\lim }\frac{1}{n^2}(2+4+6+...+2n)$

$=\underset{n\to \infty }{\lim }\frac{1}{n^2}\times 2(1+2+3+...+n)$

$=\underset{n\to \infty }{\lim }\frac{1}{n^2}\times 2\times \frac{n(n+1)}{2}=\underset{n\to \infty }{\lim }(1+\frac{1}{n})=1$

$\therefore \left(S1\right) is true.$

$\left(S2\right) : \underset{n\to \infty }{\lim }\frac{1}{n^{16}}(1^{15}+2^{15}+3^{15}+....+n^{15})=\underset{n\to \infty }{\lim }\frac{1}{n^{16}}\sum _{r=1}^n{r}^{15}$

$=\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{r=1}^n{\left(\frac{r}{n}\right)}^{15}={\int }_0^1{x}^{15}dx=|\frac{x^{16}}{16}{|}_0^1=\frac{1}{16}$

(S2) is also true.

(c) : $Let L = \underset{n\to \infty }{\lim }\left\{\right(2^{\frac{1}{2}}-2^{\frac{1}{3}}\left)\right(2^{\frac{1}{2}}-2^{\frac{1}{5}})....(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\left)\right\}$

$By Sandwich Theorem,$

${(2^{\frac{1}{2}}-2^{\frac{1}{3}})}^n<(2^{\frac{1}{2}}-2^{\frac{1}{3}})(2^{\frac{1}{2}}-2^{\frac{1}{5}})(2^{\frac{1}{2}}-2^{\frac{1}{7}})....(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}})<{(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}})}^n$

$\Rightarrow \underset{n\to \infty }{\lim } {(2^{\frac{1}{2}}-2^{\frac{1}{3}})}^n<L<\underset{n\to \infty }{\lim } {(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}})}^n$

$As, \underset{n\to \infty }{\lim } {(2^{\frac{1}{2}}-2^{\frac{1}{3}})}^n=0 and \underset{n\to \infty }{\lim } {(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}})}^n=0$

$\therefore L = 0$

(d) : $\underset{x\to 0}{\lim }\left[\right(\frac{(1-{\cos }^{2}3x)}{{\cos }^3\left(4x\right)}\left)\right(\frac{{\sin }^3\left(4x\right)}{({\log }_e{(2x+1))}^5}\left)\right]$

$=\underset{x\to 0}{\lim }[\frac{{\sin }^2\left(3x\right)}{{\cos }^{3}4x}\times \frac{{\sin }^3\left(4x\right)}{[{\log }_e{(2x+1)]}^5}]$

$=\underset{x\to 0}{\lim }\left[\frac{{\displaystyle \frac{{\sin }^2\left(3x\right)}{{\left(3x\right)}^2}}\times {\displaystyle \frac{{\sin }^3\left(4x\right)}{{\left(4x\right)}^3}}\times {\left(3x\right)}^2\times {\left(4x\right)}^3}{{\cos }^3\left(4x\right)·{\left[{\displaystyle \frac{{\log }_e(2x+1)}{2x}}\right]}^5\times {\left(2x\right)}^5}\right]$

$=\frac{9\times 64}{32}=18$

(c) : $\underset{x\to 0}{\lim }\frac{a{e}^{ax}+b\sin bx-{\displaystyle \frac{c}{2}}{e}^{-cx}+{\displaystyle \frac{c^{2}x}{2}}{e}^{-cx}}{2\sin 2x}=17$

$\Rightarrow a-\frac{c}{2}=0 \therefore a=\frac{c}{2}$

Again, Applying (L'Hospital rule)

$\underset{x\to 0}{\lim }\frac{a^2{e}^{ax}+b^2\cos bx+{\displaystyle \frac{c^2}{2}}{e}^{-cx}-{\displaystyle \frac{c^{3}x}{2}}{e}^{-cx}+{\displaystyle \frac{c^2}{2}}{e}^{-cx}}{4\cos 2x}=17$

$\Rightarrow \frac{a^2+b^2+{\displaystyle \frac{c^2}{2}}+{\displaystyle \frac{c^2}{2}}}{4}=17$

$\Rightarrow a^2+b^2+c^2=68\Rightarrow a^2+b^2+4a^2=68$

$\Rightarrow 5a^2+b^2=68$