(4)

$\text{(S1):}$ $\underset{n\to \infty }{\lim }\frac{1}{n^2}(2+4+6+\dots +2n)$

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

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

$\therefore$   $\left(S1\right)\text{ is true.}$

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

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

$\therefore$  $\left(S2\right)\text{ is also true.}$

(3)

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

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

Again, applying (L'Hospital's rule)

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

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

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

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

(3)

$\text{Let }L=\underset{t\to 0}{\lim }{(1^{\frac{{\displaystyle 1}}{{\displaystyle {\sin }^{2}t}}}+2^{\frac{{\displaystyle 1}}{{\displaystyle {\sin }^{2}t}}}+\dots +n^{\frac{{\displaystyle 1}}{{\displaystyle {\sin }^{2}t}}})}^{{\sin }^{2}t}$

$\Rightarrow L=\underset{t\to 0}{\lim }n{[{\left(\frac{{\displaystyle 1}}{{\displaystyle n}}\right)}^{cose{c}^{2}t}+{\left(\frac{{\displaystyle 2}}{{\displaystyle n}}\right)}^{cose{c}^{2}t}+ \dots +{\left(\frac{{\displaystyle n-1}}{{\displaystyle n}}\right)}^{cose{c}^{2}t}+1]}^{{\sin }^{2}t}$

$\Rightarrow L=n\left(1^{\sin \left(0\right)}\right)=n\left(1\right)=n[0<\frac{{\displaystyle k}}{{\displaystyle n}}<1,{\left(\frac{{\displaystyle k}}{{\displaystyle n}}\right)}^{\infty }=0\text{ for }1\le k<n]$

(2)

$\underset{x\to a}{\lim }\left(\right[x-5]-[2x+2\left]\right)=0$

$\Rightarrow \underset{x\to a}{\lim }\left(\right[x]-5-[2x]-2)=0 \Rightarrow \underset{x\to a}{\lim }\left(\right[x]-[2x\left]\right)=7$

$\text{Let }a\in [x_1,x+\frac{1}{2}),\text{ then }x-2x=7\Rightarrow x=-7$

$\therefore$  $a\in [-7,-6.5)$

$\text{Let }a\in (x+\frac{1}{2},x+1),\text{ then }x-(2x+1)=7$

$\Rightarrow -x=8\Rightarrow x=-8;$ $a\in [-7.5,-7)$

But limit does not exist at $a=-7.5$

$\text{Hence, }a\in (-7.5,-6.5)$

(2)

$\underset{n\to \infty }{\lim }\frac{1+2-3+4+5-6+\dots +(3n-2)+(3n-1)-3n}{\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}}$

$=\underset{n\to \infty }{\lim }\frac{(1+2+3+4+\dots +3n)-2(3+6+9+\dots +3n)}{\sqrt{2n^4+4n+3}-\sqrt{n^4+5n+4}}$

$=\underset{n\to \infty }{\lim }\frac{{\displaystyle [\frac{{\displaystyle 3n(3n+1)}}{{\displaystyle 2}}-\frac{{\displaystyle 6n(n+1)}}{{\displaystyle 2}}][\sqrt{2n^4+4n+3}+\sqrt{n^4+5n+3}]}}{(n^4-n-1)}$

$=\underset{n\to \infty }{\lim }\frac{(3n^2-3n)\times n^2\times (\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}} +\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}})}{2(n^4-n-1)}$

$=\underset{n\to \infty }{\lim }\frac{(3-\frac{3}{n})(\sqrt{2+\frac{4}{n^3}+\frac{3}{n^4}} +\sqrt{1+\frac{5}{n^3}+\frac{4}{n^4}})}{2(1-\frac{1}{n^3}-\frac{1}{n^4})}$

$=\frac{3(\sqrt{2}+1)}{2}$

(4)

$\text{Put }x=2\text{ in }{x}^2+px+q=0 \Rightarrow 4+2p+q=0 \Rightarrow -2p=(q+4)$

Now, $\underset{x\to 2p^{+}}{\lim }\frac{1-\cos (x^2-4px+q^2+8q+16)}{{(x-2p)}^4}$

$=\underset{x\to 2p^{+}}{\lim }\frac{1-\cos [x^2-2x·2p+q^2+2·q·4+4^2]}{{(x-2p)}^4}$

$=\underset{x\to 2p^{+}}{\lim }\frac{1-\cos {(x-2p)}^2}{{(x-2p)}^4}=\underset{x\to 2p^{+}}{\lim }\frac{2{\sin }^2\left(\frac{{(x-2p)}^2}{2}\right)}{{\displaystyle \frac{{(x-2p)}^4}{4}}}·\frac{1}{4}$

$=\underset{x\to 2p^{+}}{\lim }2\times 1\times \frac{1}{4}=\frac{1}{2}$      $(\because \underset{\theta \to 0}{\lim }\frac{\sin \theta }{\theta }=1)$

Now, $[1/2]=0$

(1)

$f\left(x\right)=$$\{\begin{array}{cc}\frac{x}{\left|x\right|},& x\ne 0\\ 1,& x=0\end{array}$

and $g\left(x\right)=$$\{\begin{array}{cc}\frac{\sin (x+1)}{x+1},& x\ne -1\\ 1,& x=-1\end{array}$

$h\left(x\right)=2\left[x\right]-f\left(x\right), \left[x\right]$ is the greatest integer$\le x$

$\text{LHL}$$=\underset{x\to 1^{-}}{\lim }g\left(h\right(x-1\left)\right) =\underset{k\to 0}{\lim }g\left(h\right(1-k-1\left)\right)$

$=\underset{k\to 0}{\lim }g\left[\right(2[-k]-\frac{{\displaystyle -k}}{{\displaystyle |-k|}}\left)\right]=g\left[2\right(-1)+1]=\underset{x\to 1^{-}}{\lim }g(-1)=1$

$\text{RHL}$$=\underset{x\to 1^{+}}{\lim }g\left(h\right(x-1\left)\right)=\underset{h\to 0}{\lim }g\left(h\right(1+k-1\left)\right)$

$=\underset{k\to 0}{\lim }g\left(2\right[k]-f(k\left)\right)=\underset{h\to 0}{\lim }g(0-1)=\underset{x\to 1^{+}}{\lim }g(-1)=1$

$\therefore$   $\underset{x\to 1}{\lim }g\left(h\right(x-1\left)\right)=1$

(4)

$\underset{x\to \infty }{\lim }\frac{{(\sqrt{3x+1}+\sqrt{3x-1})}^6+{(\sqrt{3x+1}-\sqrt{3x-1})}^6}{{(x+\sqrt{x^2-1})}^6+{(x-\sqrt{x^2-1})}^6}\times x^3$

$\underset{x\to \infty }{\lim }{x}^3\times \left\{\frac{{\displaystyle x^3[{(\sqrt{3+\frac{{\displaystyle 1}}{{\displaystyle x}}}+\sqrt{3-\frac{{\displaystyle 1}}{{\displaystyle x}}})}^6+{(\sqrt{3+\frac{{\displaystyle 1}}{{\displaystyle x}}}-\sqrt{3-\frac{{\displaystyle 1}}{{\displaystyle x}}})}^6]}}{{\displaystyle x^6[{(1+\sqrt{1-\frac{{\displaystyle 1}}{{\displaystyle x^2}}})}^6+{(1-\sqrt{1-\frac{{\displaystyle 1}}{{\displaystyle x^2}}})}^6]}}\right\}$

$= \frac{{{\displaystyle (}{\displaystyle \sqrt{3}}{\displaystyle +}{\displaystyle \sqrt{3}}{\displaystyle )}}^6+{(\sqrt{3}-\sqrt{3})}^6}{{{\displaystyle (}{\displaystyle 1}{\displaystyle +}{\displaystyle \sqrt{1}}{\displaystyle )}}^6+{(1-\sqrt{1})}^6}=\frac{{\left(2\sqrt{3}\right)}^6+0}{2^6+0}=\frac{2^6·3^3}{2^6}=3^3=27$