Limits and Derivatives

$$\begin{gathered} \underset{x\to 0}{\lim }cosecx\left(\sqrt{2{\cos }^{2}x+3\cos x}-\sqrt{{\cos }^{2}x+\sin x+4}\right) \\ =\underset{x\to 0}{\lim }\frac{1}{\sin x}\left[\frac{2{\cos }^{2}x+3\cos x-{\cos }^{2}x-\sin x-4}{\sqrt{2{\cos }^{2}x+3\cos x}+\sqrt{{\cos }^{2}x+\sin x+4}}\right] \\ =\underset{x\to 0}{\lim }\frac{1}{2\sqrt{5}}\left(\frac{{\cos }^{2}x+3\cos x-\sin x-4}{\sin x}\right) \\ \mathrm{using} \mathrm{L}\text{'}\mathrm{hospital} \mathrm{rule}- \\ =\frac{1}{2\sqrt{5}}\underset{x\to 0}{\lim }\left(-\frac{2\cos x\sin x-3\sin x-\cos x}{\cos x}\right) \\ =\frac{-1}{2\sqrt{5}} \end{gathered}$$

$$\begin{gathered} k^3+6k^2+11k+5 \\ =k^3+6k^2+11k+6-1 \\ =(k+1)(k+2)(k+3)-1 \end{gathered}$$

$$\begin{gathered} \underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{k^3+6k^2+11k+5}{(k+3)!} \\ =\underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{1}{k!}-\frac{1}{(k+3)!} \\ =\left(1+\frac{1}{2!}+\frac{1}{3!}+........\right) \\ -\left(\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+.........\right) \\ =1+\frac{1}{2}+\frac{1}{6}=\frac{5}{3} \end{gathered}$$

\( f(x)-6 f\left(\frac{1}{x}\right)=\frac{35}{3 x}-\frac{5}{2}\)

\( f\left(\frac{1}{x}\right)-6 f(x)=\left(\frac{35 x}{3}-\frac{5}{2}\right)\)
\( 6 f\left(\frac{1}{x}\right)-36 f(x)=\left(\frac{35 x}{3}-\frac{5}{2}\right) \times 6 \)
\( -35 f(x)=\frac{35}{3 x}-\frac{5}{2}+70 x-15 \)
\( -35 f(x)=70 x+\frac{35}{3 x}-\frac{35}{2} \)
\( f(x)=\frac{1}{2}-2 x-\frac{1}{3 x}\)
\( {lim}_{x \rightarrow 0} \frac{1}{\alpha x}+\frac{1}{2}-2 x-\frac{1}{3 x}=\beta \)
\(={lim}_{x \rightarrow 0}\left(\frac{1}{\alpha}-\frac{1}{3}\right)\frac{1}{x}+\frac{1}{2}-2 x=\beta\)
\( \alpha=3 \)
\( \beta=\frac{1}{2}\)

\(\begin{aligned}
& \lim _{x \rightarrow \infty}\left(\left(\frac{e}{e-1}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x \\
& \lim _{x \rightarrow \infty}\left(\left(\frac{1}{e-1}-\frac{e x}{(e-1)(x+1)}\right)^x\right) \\
& \alpha=e^{\lim _{x \rightarrow \infty}\left(\frac{1}{e-1}-\frac{e x}{(e-1)(x+1)}-1\right)x} \\
& \ln (\alpha)=\lim _{x \rightarrow \infty}\left(\frac{e }{(1-e)}-\frac{ex}{(1-e)(1+x)}\right) x \\
& \lim _{x \rightarrow \infty} \frac{ex}{(1-e)(1+x)}=\frac{e}{1-e} \\
& \text { Now, } \frac{\ln \alpha}{1+\ln \alpha}=\frac{\frac{e}{1-e}}{1+\frac{e}{1-e}}=e
\end{aligned}\)

Ans. (3)

Sol.

\(\begin{aligned}
& \lim _{x \rightarrow \infty} \frac{x^2\left(2-\frac{3}{x}+\frac{10}{x^2}\right)}{x^2\left(3+\frac{4}{x}-\frac{2}{x^2}\right)} \cdot\left(\frac{3 x-1}{3 x+2}\right)^{x / 2} \\
& \lim _{x \rightarrow \infty} \frac{2}{3} \times e^{\left(\frac{-3}{3 x+2}\right)\left(\frac{x}{2}\right)} \\
& =\frac{2}{3} e^{-\frac{1}{2}}=\frac{2}{3 \sqrt{e}}
\end{aligned}\)

Given: $\underset{x\to 0}{\lim }\frac{e^{(a-1)x}+2\cos bx+(c-2)e^{-x}}{x\cos x-{\log }_e(1+x)}=2,$

${\text{⇒lim}}_{x\to 0}\frac{\left(1+\left(a-1\right)x+\frac{{(a-1)}^2{x}^2}{2!}\right)+2\left(1-\frac{b^2{x}^2}{2!}\right)+\left(c-2\right)\left(1-x+\frac{x^2}{2!}\right)}{x\left(1-\frac{x^2}{2!}\right)-\left(x-\frac{x^2}{2}\dots \right)}=2$

${\text{⇒lim}}_{x\to 0}\frac{\left(1+2+c-2\right)+x\left(a-1-c+2\right)+x^2\left(\frac{{(a-1)}^2}{2}-b^2+\left(\frac{c-2}{2}\right)\right)}{\frac{x^2}{2}-\frac{x^3}{2!}+\dots }=2$

For which

$\because c+1=0\Rightarrow c=-1$

$\because a-c=-1\Rightarrow a=-2$

$\because \frac{{(\text{a}-1)}^2}{2}-{\text{b}}^2+\left(\frac{\text{c}-2}{2}\right)=1$

So, $\frac{9}{2}-{\text{b}}^2-\frac{3}{2}=1\Rightarrow {\text{ b}}^2=2$

Hence, $a^2+b^2+c^2=4+2+1=7$

${\lim }_{x\to 0}\left(f\right(2+x){)}^{3\mathrm{/}x}=(1^{\mathrm{\infty }} \text{form})$

$e^{\underset{x\to 0}{\lim }\frac{3}{x}\left(f\left(2+x\right)-1\right)}=e^{\underset{x\to 0}{\lim }3f^{\text{'}}\left(2+x\right)}$

$=e^{3f'(2)}$

$=e^{12}$

$\Rightarrow\alpha=12$

$y=4x^{3}-4x^{2}-4(12-7)x-12$

$y=4x^{3}-4x^{2}-20x-12$

$y=4\left(x^{3}-x^{2}-5x-3\right)$

$=4(x+1)^{2}(x-3)$

It meets the x-axis at two points

Given: $\underset{x\to 0}{\lim }\frac{e^{(a-1)x}+2\cos bx+(c-2)e^{-x}}{x\cos x-{\log }_e(1+x)}=2,$

${\text{⇒lim}}_{x\to 0}\frac{\left(1+\left(a-1\right)x+\frac{{(a-1)}^2{x}^2}{2!}\right)+2\left(1-\frac{b^2{x}^2}{2!}\right)+\left(c-2\right)\left(1-x+\frac{x^2}{2!}\right)}{x\left(1-\frac{x^2}{2!}\right)-\left(x-\frac{x^2}{2}\dots \right)}=2$

${\text{⇒lim}}_{x\to 0}\frac{\left(1+2+c-2\right)+x\left(a-1-c+2\right)+x^2\left(\frac{{(a-1)}^2}{2}-b^2+\left(\frac{c-2}{2}\right)\right)}{\frac{x^2}{2}-\frac{x^3}{2!}+\dots }=2$

For which

$\because c+1=0\Rightarrow c=-1$

$\because a-c=-1\Rightarrow a=-2$

$\because \frac{{(\text{a}-1)}^2}{2}-{\text{b}}^2+\left(\frac{\text{c}-2}{2}\right)=1$

So, $\frac{9}{2}-{\text{b}}^2-\frac{3}{2}=1\Rightarrow {\text{ b}}^2=2$

Hence, $a^2+b^2+c^2=4+2+1=7$

\(\lim _{x \rightarrow 0} \frac{x^2 \sin ^2 x}{x^2-\sin ^2 x}\)

Applying expansion of \(\operatorname{sin} x\)

\(=\lim _{x \rightarrow 0} \frac{x^2 \sin ^2 x}{x^2-\left(x-\frac{x^3}{3!}+\ldots\right)^2}\)

\(=\lim _{x \rightarrow 0} \frac{x^2 \sin ^2 x}{x^2-x^2+\frac{2 x^4}{3!}+ \ldots}\)

\(=\lim _{x \rightarrow 0} \frac{3!}{2}\cdot \frac{ \sin ^2 x}{x^2}\)

\(=\frac{3!}{2}=3\)

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{2{\cos }^{2}x+3\cos x-{\cos }^{2}x-\sin x-4}{\sin x(\sqrt{2{\cos }^{2}x+3\cos x}+\sqrt{{\cos }^{2}x+\sin x +4})} \\ \underset{x\to 0}{\lim }\frac{{\cos }^{2}x+3\cos x-\sin x-4}{\sin x(\sqrt{5}+\sqrt{5})} \\ L\text{'}Hospital\text{'}s Rule, \\ \underset{x\to 0}{\lim }\frac{-2\cos x \sin x - 3\sin x -\cos x}{\cos x. 2\sqrt{5}}=-\frac{1}{2\sqrt{5}} \end{gathered}$$

${\lim }_{x\to 0}\left(f\right(2+x){)}^{3\mathrm{/}x}=(1^{\mathrm{\infty }}form)$

$e^{\underset{x\to 0}{\lim }\frac{3}{x}\left(f\right(2+x)-1)}=e^{\underset{x\to 0}{\lim }3f^{\text{'}}(2+x)}$

$=e^{3f'(2)}$

$=e^{12}$

$\Rightarrow\alpha=12$

$y=4x^{3}-4x^{2}-4(12-7)x-12$

$y=4x^{3}-4x^{2}-20x-12$

$y=4\left(x^{3}-x^{2}-5x-3\right)$

$=4(x+1)^{2}(x-3)$

It meets the x-axis at two points

(B)

$$\begin{gathered} k^3+6k^2+11k+5 \\ =k^3+6k^2+11k+6-1 \\ =(k+1)(k+2)(k+3)-1 \end{gathered}$$

$$\begin{gathered} \underset{n\to \infty }{\lim }\sum _{k=1}^n\frac{k^3+6k^2+11k+5}{(k+3)!} \\ =\underset{n\to \infty }{\lim }\sum _{k=1}^n\left(\frac{1}{k!}-\frac{1}{(k+3)!}\right) \\ =\left(1+\frac{1}{2!}+\frac{1}{3!}+........\right)-\left(\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}+.........\right) \\ =1+\frac{1}{2}+\frac{1}{6}=\frac{5}{3} \end{gathered}$$

\( f(x)-6 f\left(\frac{1}{x}\right)=\frac{35}{3 x}-\frac{5}{2}\)

\( f\left(\frac{1}{x}\right)-6 f(x)=\left(\frac{35 x}{3}-\frac{5}{2}\right)\)
\( 6 f\left(\frac{1}{x}\right)-36 f(x)=\left(\frac{35 x}{3}-\frac{5}{2}\right) \times 6 \)
\( -35 f(x)=\frac{35}{3 x}-\frac{5}{2}+70 x-15 \)
\( -35 f(x)=70 x+\frac{35}{3 x}-\frac{35}{2} \)
\( f(x)=\frac{1}{2}-2 x-\frac{1}{3 x}\)
\( {lim}_{x \rightarrow 0} \frac{1}{\alpha x}+\frac{1}{2}-2 x-\frac{1}{3 x}=\beta \)
\(={lim}_{x \rightarrow 0}\left(\frac{1}{\alpha}-\frac{1}{3}\right)\frac{1}{x}+\frac{1}{2}-2 x=\beta\)
\( \alpha=3 \)
\( \beta=\frac{1}{2}\)

Given: function is continuous at \(x=0\).

So, $f\left(0\right)={\mathrm{Lim}}_{x\to 0}\frac{e^{\tan x}-e^x+\ln (\sec x+\tan x)-x}{\tan x-x}$

Applying L'hopital rule,

$\Rightarrow f\left(0\right)={\mathrm{Lim}}_{x\to 0}\frac{e^{\tan x}\cdot {\sec }^{2}x-e^x+\sec x-1}{{\sec }^{2}x-1}$

$\Rightarrow f\left(0\right)={\mathrm{Lim}}_{x\to 0}\frac{e^{\tan x}\left({\sec }^{2}x-1\right)+\left(e^{\tan x}-e^x\right)+\sec x-1}{{\tan }^{2}x}$

$$\begin{gathered} \Rightarrow f\left(0\right)={\mathrm{Lim}}_{x\to 0}\left(e^{\tan x}+\frac{e^x\left(e^{\tan x-x}-1\right)}{{\tan }^{2}x}+\frac{1}{\sec x+1}\right) \\ ={\mathrm{Lim}}_{x\to 0}\left(e^{\tan x}+\frac{e^x\left(e^{\tan x-x}-1\right)\left(\tan x-x\right)}{{\displaystyle {\tan }^{2}x}\left(\tan x-x\right)}+\frac{1}{\sec x+1}\right) \end{gathered}$$

since

$$\begin{gathered} {\mathrm{Lim}}_{x\to 0}\frac{\tan x-x}{{\tan }^{2}x}={\mathrm{Lim}}_{x\to 0}\frac{{\sec }^{2}x-1}{2\tan x {\sec }^{2}x} \\ ={\mathrm{Lim}}_{x\to 0}\frac{\tan x}{2 {\sec }^{2}x}=0 \end{gathered}$$

Therefore

$f\left(0\right)=1+0+\frac{1}{2}=\frac{3}{2}$

Ans. (3)

Sol.

\(\begin{aligned}
& \lim _{x \rightarrow \infty} \frac{x^2\left(2-\frac{3}{x}+\frac{10}{x^2}\right)}{x^2\left(3+\frac{4}{x}-\frac{2}{x^2}\right)} \cdot\left(\frac{3 x-1}{3 x+2}\right)^{x / 2} \\
& \lim _{x \rightarrow \infty} \frac{2}{3} \times e^{\left(\frac{-3}{3 x+2}\right)\left(\frac{x}{2}\right)} \\
& =\frac{2}{3} e^{-\frac{1}{2}}=\frac{2}{3 \sqrt{e}}
\end{aligned}\)

${\lim }_{x\to 0}\left(f\right(2+x){)}^{3\mathrm{/}x}=(1^{\mathrm{\infty }}form)$

$e^{\underset{x\to 0}{\lim }\frac{3}{x}\left(f\right(2+x)-1)}=e^{\underset{x\to 0}{\lim }3f^{\text{'}}(2+x)}$

$=e^{3f'(2)}$

$=e^{12}$

$\Rightarrow\alpha=12$

$y=4x^{3}-4x^{2}-4(12-7)x-12$

$y=4x^{3}-4x^{2}-20x-12$

$y=4\left(x^{3}-x^{2}-5x-3\right)$

$=4(x+1)^{2}(x-3)$

It meets the x-axis at two points

$$\begin{gathered} \underset{x\to 0}{\lim }cosecx\left(\sqrt{2{\cos }^{2}x+3\cos x}-\sqrt{{\cos }^{2}x+\sin x+4}\right) \\ =\underset{x\to 0}{\lim }\frac{1}{\sin x}\left[\frac{2{\cos }^{2}x+3\cos x-{\cos }^{2}x-\sin x-4}{\sqrt{2{\cos }^{2}x+3\cos x}+\sqrt{{\cos }^{2}x+\sin x+4}}\right] \\ =\underset{x\to 0}{\lim }\frac{1}{2\sqrt{5}}\left(\frac{{\cos }^{2}x+3\cos x-\sin x-4}{\sin x}\right) \\ \mathrm{using} \mathrm{L}\text{'}\mathrm{hospital} \mathrm{rule}- \\ =\frac{1}{2\sqrt{5}}\underset{x\to 0}{\lim }\left(-\frac{2\cos x\sin x-3\sin x-\cos x}{\cos x}\right) \\ =\frac{-1}{2\sqrt{5}} \end{gathered}$$

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{2{\cos }^{2}x+3\cos x-{\cos }^{2}x-\sin x-4}{\sin x(\sqrt{2{\cos }^{2}x+3\cos x}+\sqrt{{\cos }^{2}x+\sin x +4})} \\ \underset{x\to 0}{\lim }\frac{{\cos }^{2}x+3\cos x-\sin x-4}{\sin x(\sqrt{5}+\sqrt{5})} \\ L\text{'}Hospital\text{'}s Rule, \\ \underset{x\to 0}{\lim }\frac{-2\cos x \sin x - 3\sin x -\cos x}{\cos x. 2\sqrt{5}}=-\frac{1}{2\sqrt{5}} \end{gathered}$$

\(\begin{aligned}
& \lim _{x \rightarrow \infty}\left(\left(\frac{e}{e-1}\right)\left(\frac{1}{e}-\frac{x}{1+x}\right)\right)^x \\
& \lim _{x \rightarrow \infty}\left(\left(\frac{1}{e-1}-\frac{e x}{(e-1)(x+1)}\right)^x\right) \\
& \alpha=e^{\lim _{x \rightarrow \infty}\left(\frac{1}{e-1}-\frac{e x}{(e-1)(x+1)}-1\right)x} \\
& \ln (\alpha)=\lim _{x \rightarrow \infty}\left(\frac{e }{(1-e)}-\frac{ex}{(1-e)(1+x)}\right) x \\
& \lim _{x \rightarrow \infty} \frac{ex}{(1-e)(1+x)}=\frac{e}{1-e} \\
& \text { Now, } \frac{\ln \alpha}{1+\ln \alpha}=\frac{\frac{e}{1-e}}{1+\frac{e}{1-e}}=e
\end{aligned}\)

\(\lim _{x \rightarrow 2} \frac{\tan (x-2)}{x-2} \cdot \frac{r x^2+(p-2) x-2 p}{x-2}=5\)

Since \(\lim _{x \rightarrow 2} \frac{\tan (x-2)}{x-2}=1\), we need

\(\lim _{x \rightarrow 2} \frac{r x^2+(p-2) x-2 p}{x-2}=5\)

For the limit to be finite,

\(4 r+2(p-2)-2 p=0 \Rightarrow 4 r-4=0 \Rightarrow r=1\)

Now \(x^2+(p-2) x-2 p=(x-2)(x+p)\)

So \(\lim _{x \rightarrow 2}(x+p)=5 \Rightarrow 2+p=5 \Rightarrow p=3\)

The quadratic equation becomes \(x^2-3 x+q=0\)

For real roots, \(\Delta \geq 0 \Rightarrow 9-4 q \geq 0 \Rightarrow q \leq \frac{9}{4}\)

For both roots to lie in \((0,2)\), \(f(0)=q>0\) and \(f(2)>0\)

\(f(2)=4-6+q=q-2>0 \Rightarrow q>2\)

So, \(q \in\left(2, \frac{9}{4}\right]\)

Thus, \(\alpha=2, \beta=\frac{9}{4}\)

\(4(\alpha+\beta)=4\left(2+\frac{9}{4}\right)=17\)