Differentiation (Part-38) | S N De

 $\,5.\

,$ Find the values of the following limits :

$\,(i)\,\lim_{n \to \infty}\frac{1+2+3+\cdots+n}{n^2}$

Sol. $~~\lim_{n \to \infty}\frac{1+2+3+\cdots+n}{n^2}\\=\lim_{n \to \infty} \frac{n(n+1)}{2n^2}\\=\lim_{n \to \infty}\frac{n^2\left(1+\frac 1n\right)}{2n^2}\\=\frac 12 \lim_{n \to \infty} \left(1+\frac 1n\right)\\=\frac 12 \times (1+0)\\=\frac 12\,\,\,\text{(ans.)}$

$\,(ii)~~\lim_{n \to \infty}\frac{1^2+2^2+3^2+\cdots+n^2}{n^3}$

Sol. $\,~~\lim_{n \to \infty}\frac{1^2+2^2+3^2+\cdots+n^2}{n^3}\\=\lim_{n \to \infty}\frac{n(n+1)(2n+1)}{6n^3}\\=\lim_{n \to \infty}\frac{n^3\left(1+\frac 1n\right)\left(2+\frac 1n\right)}{6n^3}\\=\frac 16 \times \lim_{n \to \infty} \left(1+\frac 1n\right)\left(2+\frac 1n\right)\\=\frac 16 \times 1\times 2\\=\frac 13\,\,\text{(ans.)}$

$\,(iii)~~\lim_{n \to \infty}\left(1+\frac 12+\frac {1}{2^2}+\frac{1}{2^3}+\cdots\text{to n terms}\right)$

Sol. $\,~~\lim_{n \to \infty}\left(1+\frac 12+\frac {1}{2^2}+\frac{1}{2^3}+\cdots\text{to n terms}\right)\\=\lim_{n \to \infty}~\left[ 1 \cdot \frac{1-\left(\frac 12\right)^n}{1-\frac 12}\right]\\=1 \cdot \frac{1-0}{1-\frac 12}\,\,[\because \left(\frac 12\right)^n \to 0 ~~\text{as}~~n \to \infty] \\=\frac{1}{1/2}\\=2\,\,\text{(ans.)}$

$\,(iv)\,~~\lim_{n \to \infty}  \left(\frac 13+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^n}\right)$

Sol. $\,\,~~\lim_{n \to \infty}  \left(\frac 13+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^n}\right)\\=\lim_{n \to \infty}  \left[\frac 13\cdot \frac{1- \left(\frac 13\right)^n}{1-\frac 13}\right]\\=\frac 13 \times \frac{1-0}{\frac 23} \,\,[\because \left(\frac 13\right)^n \to 0 ~~\text{as}~~n \to \infty] \\=\frac 13 \times \frac 32\\=\frac 12\,\,\text{(ans.)}$

$\,(v)\,\lim_{n \to \infty} \frac{1+2+3+\cdots\text{to (3n+2) terms}}{(n+1)^2}$

Sol. $\,\,\lim_{n \to \infty} \frac{1+2+3+\cdots\text{to (3n+2) terms}}{(n+1)^2}\\=\lim_{n \to \infty} \left[\frac{\frac{(3n+2)(3n+2+1)}{2}}{(n+1)^2}\right]\\=\lim_{n \to \infty}\left[ \frac 12 \cdot \frac{3n \left(1+\frac{2}{3n}\right) \cdot 3n  \left(1+\frac 1n\right)}{n^2 \left(1+\frac 1n\right)^2}\right]\\=\lim_{n \to \infty}\left[ \frac 92 \cdot \frac{ \left(1+\frac{2}{3n}\right) \cdot  \left(1+\frac 1n\right)}{ \left(1+\frac 1n\right)^2}\right]\\=\frac 92 \cdot \frac{(1+0)(1+0)}{(1+0)^2}\\=\frac 92\,\,\text{(ans.)}$

$\,(vi)\,\lim_{n \to \infty} \frac{1\cdot2+2\cdot3+3\cdot4+\cdots +n(n+1)}{n^3}$

Sol. Let us consider the series :

$\,S=1\cdot2+2\cdot3+3\cdot4+\cdots +n(n+1)\rightarrow(1)$

Now, $\,n\,$th term of $\,(1),\,$ is :

$t_n=n(n+1)=n^2+n\\ \therefore S=\sum{t_n}\\=\sum{n^2}+\sum{n}\\=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}\\=\frac 16[n(n+1)(2n+1)+3n(n+1)]\rightarrow(2)$

So, $\,\lim_{n \to \infty} \frac{1\cdot2+2\cdot3+3\cdot4+\cdots +n(n+1)}{n^3}\\=\lim_{n \to \infty} \left[\frac 16 \cdot\frac{n(n+1)(2n+1)+3n(n+1)}{n^3}\right]\\=\frac 16 \cdot \lim_{n \to \infty} \frac{n^3 \left(1+\frac 1n\right) \left(2+\frac 1n\right)+3n^2 \left(1+\frac 1n\right)}{n^3}\\=\frac 16 \lim_{n \to \infty} \left[\left(1+\frac 1n\right) \left(2+\frac 1n\right)+\frac 3n  \left(1+\frac 1n\right)\right]\\=\frac 16[(1+0)(2+0)+0]\\=\frac 16 \times (1+2)\\=\frac 16 \times 2\\=\frac 13\,\,\text{(ans.)}$

$\,6(i)\,\,$ Prove that $\,\lim_{x \to \infty} \frac{a\cdot e^x+b\cdot e^{-x}}{c\cdot e^x+d\cdot e^{-x}}=\frac ac~~~[2<e<3, c \neq 0]$

Sol. $~~~\lim_{x \to \infty} \frac{a\cdot e^x+b\cdot e^{-x}}{c\cdot e^x+d\cdot e^{-x}}\\=\lim_{x \to \infty} \frac{e^x(a+b \cdot e^{-2x})}{e^x(c+d \cdot e^{-2x})}\\=\lim_{x \to \infty} \frac{(a+b \cdot e^{-2x})}{(c+d \cdot e^{-2x})}\\=\frac{a+0}{c+0}\\~~~[\text{as}~~ x \to \infty, e^{-2x}\to 0,~~\text{since}~~ e^{-2x}<1]\\=\frac ac\,\,\text{(proved)}$

$\,(ii)\,$ Show that $\,\lim_{x \to -\infty} \frac{a\cdot e^x+b\cdot e^{-x}}{c\cdot e^x+d\cdot e^{-x}}=\frac bd~~~[2<e<3, d \neq 0]$

Sol. Let $\,x=-z\,$ so that as $\,x \to -\infty,z \to \infty.$

So, $\,\lim_{x \to -\infty} \frac{a\cdot e^x+b\cdot e^{-x}}{c\cdot e^x+d\cdot e^{-x}}\\=\lim_{z \to \infty} \frac{a \cdot e^{-z}+b \cdot e^z}{c \cdot e^{-z}+d \cdot e^z}\\=\lim_{z \to \infty} \frac{e^z(a  \cdot e^{-2z}+b)}{e^z(c \cdot e^{-2z}+d)}\\=\lim_{z \to \infty}\frac{(a  \cdot e^{-2z}+b)}{(c  \cdot e^{-2z}+d)}\\=\frac{0+b}{0+d}\\~~~[\text{as}~~ z \to \infty, e^{-2z}\to 0,~~\text{since}~~ e^{-2z}<1]\\=\frac bd\,\,\text{(proved)}$

$\,7.\,$ Do the following limits exist ?

$\,(i)\, \lim_{x \to \pi} \frac{1}{x-\pi}$

Sol. As $\,x \to \pi,\, \pi-x \to 0.$

$ \therefore ~~\lim_{x \to \pi} \frac{1}{x-\pi}\,$ does not exist.

$\,(ii)\, \lim_{x \to 0}\cos \frac 1x$

Sol. Let $\,x=\frac{1}{2n\pi+\theta}~~(-\pi <\theta\leq \pi).$ 

Now, as $\,n \to \infty, x \to 0.$

So, $\,\,L= \lim_{x \to 0}\cos \frac 1x\\=\lim_{n \to \infty} \cos(2n\pi+\theta)\\=\lim_{n \to \infty} \cos\theta\\=\cos\theta \rightarrow(1)$

Hence, from $\,(1),\,$ it is clear that $\,L\,$ can be different  for different  values of $\,\,\theta.$ 

So, $\,\lim_{x \to 0}\cos \frac 1x\,$ does not exist.

$\,(iii)\,\lim_{x \to 3}[x^2+\sqrt{3-x}]$

Sol. For $\,x \to 3+,\,[x^2+\sqrt{3-x}]\,$ gives imaginary value.

Again, $\,\lim_{x \to 3-}[x^2+\sqrt{3-x}]\\=3^2+0\\=9.$

So, $\,\lim_{x \to 3+}[x^2+\sqrt{3-x}]\\ \neq \lim_{x \to 3-}[x^2+\sqrt{3-x}].$

Hence, $\,\,\lim_{x \to 3}[x^2+\sqrt{3-x}]\,\,$ does not exist.

$\,(iv)\,\lim_{x \to 2}\frac{1}{3+e^{\frac{1}{x-2}}}\,~~~[2<e<3]$

Sol. As $\,x \to 2+,~~e^{\frac{1}{x-2}} \to \infty. \\ \therefore \lim_{x \to 2+}\frac{1}{3+e^{\frac{1}{x-2}}}=0.$

Again, as $\,x \to 2-,~~\frac{1}{x-2}\to -\infty.$

So, $~~e^{\frac{1}{x-2}} \to 0,\,\, \text{for}~~x \to 2-.$

$ \therefore \lim_{x \to 2-}\frac{1}{3+e^{\frac{1}{x-2}}}=\frac{1}{3+0}=\frac 13.$

Hence, $\,\,\lim_{x \to 2+}\frac{1}{3+e^{\frac{1}{x-2}}}  \neq \lim_{x \to 2-}\frac{1}{3+e^{\frac{1}{x-2}}}.$

$\,\therefore \lim_{x \to 2}\frac{1}{3+e^{\frac{1}{x-2}}}\,\,$ does not exist.