Differentiation (Part-38) | S N De

$1.$ Draw the graph of the function $y=\frac{x^2}{x}$ and from the graph evaluate $\lim_{x \to 0}\frac{x^2}{x}.$

Solution

From the graph, using the definition of limit, we can evaluate the limit of the given function i.e., $f(x)=\frac{x^2}{x}.$

Let us denote Left hand limit by L.H.L. and Right hand limit by R.H.L.

$$\text{L.H.L.} = \lim_{h \to 0} f(0-h) = \lim_{h \to 0}\frac{(0-h)^2}{(0-h)} = \lim_{h \to 0}\frac{h^2}{-h} = \lim_{h \to 0} (-h) = 0.$$

$$\text{Similarly, R.H.L.} = \lim_{h \to 0}f(0+h) = \lim_{h \to 0}\frac{(0+h)^2}{(0+h)} = \lim_{h \to 0}\frac{h^2}{h} = \lim_{h \to 0} (h) = 0.$$

$\therefore \text{L.H.L.} = \text{R.H.L.}$ So, the limit exists and its value is $0.$

$2.$ Sketch the graph of: $$f(x) = \begin{cases} 1, & \text{for } x > 0 \\ x^2, & \text{for } x \le 0 \end{cases}$$

Hint: Follow S N Dey worked out problem: [EXAMPLE 7]

3. Proof-Based Limit Problems

$3(i)$ Prove: $$\lim_{x\to 4} \frac{3-\sqrt{5+x}}{1-\sqrt{5-x}} = -\frac{1}{3}$$

Solution

$$\lim_{x\to 4} \frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}$$ $$= \lim_{x \to 4}\frac{(3-\sqrt{5+x})(1+\sqrt{5-x})}{(1-\sqrt{5-x})(1+\sqrt{5-x})}$$ $$= \lim_{x \to 4} \frac{(3-\sqrt{5+x})(3+\sqrt{5+x})(1+\sqrt{5-x})}{[1^2-(\sqrt{5-x})^2](3+\sqrt{5+x})}$$ $$= \lim_{x\to 4} \frac{(3^2-5-x)(1+\sqrt{5-x})}{(1-5+x)(3+\sqrt{5+x})}$$ $$= \lim_{x\to 4}\frac{-(x-4)(1+\sqrt{5-x})}{(x-4)(3+\sqrt{5+x})}$$ $$= -\frac{1+\sqrt{5-4}}{3+\sqrt{5+4}} = -\frac{1+\sqrt{1}}{3+\sqrt{9}} = -\frac{1+1}{3+3} = -\frac{2}{6} = -\frac{1}{3} \quad \text{(Proved)}$$

$3(ii)$ Prove: $$\lim_{x \to 1}\frac{x^2-\sqrt{x}}{\sqrt{x}-1} = 3$$

Solution

$$\lim_{x \to 1}\frac{x^2-\sqrt{x}}{\sqrt{x}-1}$$ $$= \lim_{x \to 1} \frac{(x^2-\sqrt{x})(\sqrt{x}+1)(x^2+\sqrt{x})}{(\sqrt{x}-1)(\sqrt{x}+1)(x^2+\sqrt{x})}$$ $$= \lim_{x \to 1}\frac{[(x^2)^2-(\sqrt{x})^2](\sqrt{x}+1)}{[(\sqrt{x})^2-(1)^2](x^2+\sqrt{x})}$$ $$= \lim_{x \to 1}\frac{(x^4-x)(\sqrt{x}+1)}{(x-1)(x^2+\sqrt{x})}$$ $$= \lim_{x \to 1} \frac{x(x^3-1)(\sqrt{x}+1)}{(x-1)(x^2+\sqrt{x})}$$ $$= \lim_{x \to 1}\frac{x(x-1)(x^2+x+1)(\sqrt{x}+1)}{(x-1)(x^2+\sqrt{x})}$$ $$= \lim_{x \to 1}\frac{x(x^2+x+1)(\sqrt{x}+1)}{x^2+\sqrt{x}}$$ $$= \frac{1(1^2+1+1)(\sqrt{1} +1)}{1^2+\sqrt{1}} = \frac{3 \times 2}{2} = 3 \quad \text{(Proved)}$$

$3(iii)$ Prove: $$\lim_{x \to 0} \frac{1-\cos x \sqrt{\cos 2x}}{x^2} = \frac{3}{2}$$

Solution

$$\lim_{x \to 0} \frac{1-\cos x \sqrt{\cos 2x}}{x^2}$$ $$= \lim_{x \to 0} \frac{(1-\cos x \sqrt{\cos 2x})(1+\cos x\sqrt{\cos 2x})}{x^2 (1+\cos x\sqrt{\cos 2x})}$$ $$= \lim_{x \to 0}\frac{[1-\cos^2x(\cos 2x)]}{x^2(1+\cos x\sqrt{\cos 2x})}$$ $$= \lim_{x \to 0}\frac{1-(1-\sin^2x)\cos 2x}{x^2(1+\cos x\sqrt{\cos 2x})}$$ $$= \lim_{x \to 0}\frac{1-\cos 2x+\sin^2x\cos 2x}{x^2(1+\cos x\sqrt{\cos 2x})}$$ $$= \lim_{x \to 0}\frac{2\sin^2x+\sin^2x\cos 2x}{x^2(1+\cos x\sqrt{\cos 2x})}$$ $$= \lim_{x \to 0}\frac{\sin^2x(2+\cos 2x)}{x^2(1+\cos x\sqrt{\cos 2x})}$$ $$= \lim_{x \to 0} \frac{\sin^2x}{x^2} \cdot \lim_{x \to 0}\frac{2+\cos 2x}{(1+\cos x\sqrt{\cos 2x})}$$ $$= 1 \cdot \frac{2+1}{1+1 \cdot 1} = \frac{3}{2} \quad \text{(Proved)}$$

4. Limits at Infinity

$4(i)$ Prove: $$\lim_{n \to \infty} (\sqrt{n+1}-\sqrt{n}) = 0$$

Solution

$$\lim_{n \to \infty} (\sqrt{n+1}-\sqrt{n})$$ $$= \lim_{n \to \infty}\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{(\sqrt{n+1}+\sqrt{n})}$$ $$= \lim_{n \to \infty}\frac{(\sqrt{n+1})^2-(\sqrt{n})^2}{(\sqrt{n+1}+\sqrt{n})}$$ $$= \lim_{n \to \infty}\frac{n+1-n}{(\sqrt{n+1}+\sqrt{n})}$$ $$= \lim_{n \to \infty}\frac{1}{(\sqrt{n+1}+\sqrt{n})}$$ $$= \lim_{n \to \infty}\frac{1/\sqrt{n}}{\sqrt{1+\frac{1}{n}}+1} = 0 \quad \left[\because \frac{1}{\sqrt{n}} \to 0 \text{ as } n \to \infty\right] \quad \text{(Proved)}$$

$4(ii)$ Prove: $$\lim_{x \to \infty} \sqrt{x}[\sqrt{x+2}-\sqrt{x}] = 1$$

Solution

$$\lim_{x \to \infty} \sqrt{x}[\sqrt{x+2}-\sqrt{x}]$$ $$= \lim_{x \to \infty} \frac{x[\sqrt{x+2}-\sqrt{x}][\sqrt{x+2}+\sqrt{x}]}{\sqrt{x}(\sqrt{x+2}+\sqrt{x})}$$ $$= \lim_{x \to \infty}\frac{x[(\sqrt{x+2})^2-(\sqrt{x})^2]}{\sqrt{x}(\sqrt{x+2}+\sqrt{x})}$$ $$= \lim_{x \to \infty}\frac{x(x+2-x)}{\sqrt{x}(\sqrt{x+2}+\sqrt{x})}$$ $$= \lim_{x \to \infty}\frac{2x}{x\left(\sqrt{1+\frac{2}{x}}+1\right)} = \frac{2}{1+1} \quad \left[\because \frac{1}{x} \to 0 \text{ as } x\to \infty\right] = 1 \quad \text{(Proved)}$$

$4(iii)$ Prove: $$\lim_{x \to \infty} [\sqrt{1+x+x^2}-x] = \frac{1}{2}$$

Solution

$$\lim_{x \to \infty}[\sqrt{1+x+x^2}-x]$$ $$= \lim_{x \to \infty}\frac{(\sqrt{1+x+x^2}-x)(\sqrt{1+x+x^2}+x)}{(\sqrt{1+x+x^2}+x)}$$ $$= \lim_{x \to \infty}\frac{(\sqrt{1+x+x^2})^2-(x)^2}{(\sqrt{1+x+x^2}+x)}$$ $$= \lim_{x \to \infty}\frac{1+x+x^2-x^2}{(\sqrt{1+x+x^2}+x)}$$ $$= \lim_{x \to \infty}\frac{1+x}{(\sqrt{1+x+x^2}+x)}$$ $$= \lim_{x \to \infty}\frac{x\left(\frac{1}{x}+1\right)}{x\left(\sqrt{\frac{1}{x^2}+\frac{1}{x}+1}+1\right)}$$ $$= \lim_{x \to 1}\frac{\left(\frac{1}{x}+1\right)}{\left(\sqrt{\frac{1}{x^2}+\frac{1}{x}+1}+1\right)} = \frac{1}{1+1} = \frac{1}{2} \quad \text{(Proved)}$$

$4(iv)$ Prove: $$\lim_{n \to \infty} [\sqrt[3]{n+1}-\sqrt[3]{n}] = 0$$

Solution

$$\lim_{n \to \infty} [\sqrt[3]{n+1}-\sqrt[3]{n}]$$ $$= \lim_{n \to \infty }\frac{[\sqrt[3]{n+1}-\sqrt[3]{n}] \times [(n+1)^{\frac{2}{3}}+n^{\frac{1}{3}}(n+1)^{\frac{1}{3}}+n^{\frac{2}{3}}]}{[(n+1)^{\frac{2}{3}}+n^{\frac{1}{3}}(n+1)^{\frac{1}{3}}+n^{\frac{2}{3}}]}$$ $$= \lim_{n \to \infty} \frac{\left(\sqrt[3]{n+1}\right)^3-\left(\sqrt[3]{n}\right)^3}{(n+1)^{\frac{2}{3}}+n^{\frac{1}{3}}(n+1)^{\frac{1}{3}}+n^{\frac{2}{3}}}$$ $$= \lim_{n \to \infty}\frac{n+1-n}{(n+1)^{\frac{2}{3}}+n^{\frac{1}{3}}(n+1)^{\frac{1}{3}}+n^{\frac{2}{3}}}$$ $$= \lim_{n \to \infty} \frac{1}{(n+1)^{\frac{2}{3}}+n^{\frac{1}{3}}(n+1)^{\frac{1}{3}}+n^{\frac{2}{3}}} = 0 \quad [*] \quad \text{(Proved)}$$

Note [*]: As $n \to \infty$, the denominator approaches $\infty$, causing the fraction to evaluate to $0$.