Differentiation (Part-38) | S N De

5. Evaluation of Infinite Limits and Series

$5(i)$ Find the value of the limit: $$\lim_{n \to \infty}\frac{1+2+3+\cdots+n}{n^2}$$

Solution

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

$5(ii)$ Find the value of the limit: $$\lim_{n \to \infty}\frac{1^2+2^2+3^2+\cdots+n^2}{n^3}$$

Solution

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

$5(iii)$ Find the value of the limit: $$\lim_{n \to \infty}\left(1+\frac 12+\frac {1}{2^2}+\frac{1}{2^3}+\cdots \text{ to } n \text{ terms}\right)$$

Solution

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

$5(iv)$ Find the value of the limit: $$\lim_{n \to \infty} \left(\frac 13+\frac{1}{3^2}+\frac{1}{3^3}+\cdots+\frac{1}{3^n}\right)$$

Solution

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

$5(v)$ Find the value of the limit: $$\lim_{n \to \infty} \frac{1+2+3+\cdots \text{ to } (3n+2) \text{ terms}}{(n+1)^2}$$

Solution

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

$5(vi)$ Find the value of the limit: $$\lim_{n \to \infty} \frac{1\cdot2+2\cdot3+3\cdot4+\cdots +n(n+1)}{n^3}$$

Solution

Let us consider the series:

$$S = 1\cdot2+2\cdot3+3\cdot4+\cdots +n(n+1) \quad \dots (1)$$

Now, the $n$-th term of $(1)$ is $t_n = n(n+1) = n^2+n$.

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

Evaluating the limit:

$$\lim_{n \to \infty} \frac{S}{n^3} = \lim_{n \to \infty} \left[\frac 16 \cdot\frac{n(n+1)(2n+1)+3n(n+1)}{n^3}\right]$$ $$= \frac 16 \cdot \lim_{n \to \infty} \frac{n^3 \left(1+\frac 1n\right) \left(2+\frac 1n\right)+3n^2 \left(1+\frac 1n\right)}{n^3}$$ $$= \frac 16 \lim_{n \to \infty} \left[\left(1+\frac 1n\right) \left(2+\frac 1n\right)+\frac 3n \left(1+\frac 1n\right)\right]$$ $$= \frac 16[(1+0)(2+0)+0] = \frac 16 \times 2 = \frac 13 \quad \text{(Ans.)}$$

6. Limits Involving Exponential Functions

$6(i)$ 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 \quad [2

Solution

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

$6(ii)$ 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 \quad [2

Solution

Let $x = -z$. As $x \to -\infty$, $z \to \infty$.

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

7. Existence of Limits

$7(i)$ Check if the following limit exists: $$\lim_{x \to \pi} \frac{1}{x-\pi}$$

Solution

As $x \to \pi$, $(x-\pi) \to 0$.

Conclusion: Limit does not exist

$7(ii)$ Check if the following limit exists: $$\lim_{x \to 0}\cos \frac 1x$$

Solution

Let $x = \frac{1}{2n\pi+\theta}$ where $-\pi < \theta \le \pi$.

As $n \to \infty$, $x \to 0$.

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

From equation $(1)$, it is clear that $L$ can take different values for different values of $\theta$.

Conclusion: Limit does not exist

$7(iii)$ Check if the following limit exists: $$\lim_{x \to 3}\left[x^2+\sqrt{3-x}\right]$$

Solution

For $x \to 3^+$, $\left[x^2+\sqrt{3-x}\right]$ produces imaginary values as $(3-x) < 0$.

Evaluating Left Hand Limit:

$$\lim_{x \to 3-}\left[x^2+\sqrt{3-x}\right] = 3^2+0 = 9$$

Since $\lim_{x \to 3+}\left[x^2+\sqrt{3-x}\right] \neq \lim_{x \to 3-}\left[x^2+\sqrt{3-x}\right]$,

Conclusion: Limit does not exist

$7(iv)$ Check if the following limit exists: $$\lim_{x \to 2}\frac{1}{3+e^{\frac{1}{x-2}}} \quad [2

Solution

Right Hand Limit (R.H.L.):

As $x \to 2^+$, $\frac{1}{x-2} \to +\infty \implies e^{\frac{1}{x-2}} \to \infty$.

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

Left Hand Limit (L.H.L.):

As $x \to 2^-$, $\frac{1}{x-2} \to -\infty \implies e^{\frac{1}{x-2}} \to 0$.

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

Since $\text{R.H.L.} \neq \text{L.H.L.}$,

Conclusion: Limit does not exist