Differentiation (Part-38) | S N De

6–8. Limits Involving Double Rationalization and Difference Quotients

$6.$ Evaluate: $$\lim_{x \to 0} \frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}$$

Solution

Multiplying both numerator and denominator by their respective conjugate factors:

$$\lim_{x \to 0} \frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}$$ $$= \lim_{x \to 0}\frac{(\sqrt{1+x^2}-\sqrt{1+x})(\sqrt{1+x^2}+\sqrt{1+x})(\sqrt{1+x^3}+\sqrt{1+x})}{(\sqrt{1+x^3}-\sqrt{1+x})(\sqrt{1+x^3}+\sqrt{1+x})(\sqrt{1+x^2}+\sqrt{1+x})}$$ $$= \lim_{x \to 0}\frac{\{(1+x^2)-(1+x)\}(\sqrt{1+x^3}+\sqrt{1+x})}{\{(1+x^3)-(1+x)\}(\sqrt{1+x^2}+\sqrt{1+x})}$$ $$= \lim_{x \to 0}\frac{(x^2-x)(\sqrt{1+x^3}+\sqrt{1+x})}{(x^3-x)(\sqrt{1+x^2}+\sqrt{1+x})}$$ $$= \lim_{x \to 0}\frac{x(x-1)(\sqrt{1+x^3}+\sqrt{1+x})}{x(x-1)(x+1)(\sqrt{1+x^2}+\sqrt{1+x})}$$ $$= \lim_{x \to 0}\frac{\sqrt{1+x^3}+\sqrt{1+x}}{(x+1)(\sqrt{1+x^2}+\sqrt{1+x})}$$ $$= \frac{\sqrt{1+0^3}+\sqrt{1+0}}{(0+1)(\sqrt{1+0^2}+\sqrt{1+0})} = \frac{1+1}{1 \times (1+1)} = 1 \quad \text{(Ans.)}$$

$7.$ Evaluate: $$\lim_{x \to a}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}} \quad (a \neq 0)$$

Solution

Rationalizing both numerator and denominator simultaneously:

$$\lim_{x \to a}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}$$ $$= \lim_{x \to a}\frac{(\sqrt{a+2x}-\sqrt{3x})(\sqrt{a+2x}+\sqrt{3x})(\sqrt{3a+x}+2\sqrt{x})}{(\sqrt{3a+x}-2\sqrt{x})(\sqrt{3a+x}+2\sqrt{x})(\sqrt{a+2x}+\sqrt{3x})}$$ $$= \lim_{x \to a}\frac{[(a+2x)-3x](\sqrt{3a+x}+2\sqrt{x})}{[(3a+x)-4x](\sqrt{a+2x}+\sqrt{3x})}$$ $$= \lim_{x \to a}\frac{(a-x)(\sqrt{3a+x}+2\sqrt{x})}{3(a-x)(\sqrt{a+2x}+\sqrt{3x})}$$ $$= \lim_{x \to a}\frac{\sqrt{3a+x}+2\sqrt{x}}{3(\sqrt{a+2x}+\sqrt{3x})}$$ $$= \frac{\sqrt{3a+a}+2\sqrt{a}}{3(\sqrt{a+2a}+\sqrt{3a})} = \frac{\sqrt{4a}+2\sqrt{a}}{3(\sqrt{3a}+\sqrt{3a})} = \frac{2\sqrt{a}+2\sqrt{a}}{3 \times 2\sqrt{3a}}$$ $$= \frac{4\sqrt{a}}{6\sqrt{3}\sqrt{a}} = \frac{2}{3\sqrt{3}} \quad \text{(Ans.)}$$

$8.$ If $G(x) = -\sqrt{25-x^2}$, find the value of: $$\lim_{x \to 1}\frac{G(x)-G(1)}{x-1}$$

Solution

Evaluating $G(1) = -\sqrt{25-1^2} = -\sqrt{24}$:

$$\lim_{x \to 1}\frac{G(x)-G(1)}{x-1} = \lim_{x \to 1}\frac{-\sqrt{25-x^2} - (-\sqrt{24})}{x-1}$$ $$= \lim_{x \to 1}\frac{\sqrt{24}-\sqrt{25-x^2}}{x-1}$$

Rationalizing the numerator:

$$= \lim_{x \to 1}\frac{(\sqrt{24}-\sqrt{25-x^2})(\sqrt{24}+\sqrt{25-x^2})}{(x-1)(\sqrt{24}+\sqrt{25-x^2})}$$ $$= \lim_{x \to 1}\frac{24-(25-x^2)}{(x-1)(\sqrt{24}+\sqrt{25-x^2})}$$ $$= \lim_{x \to 1}\frac{x^2-1}{(x-1)(\sqrt{24}+\sqrt{25-x^2})}$$ $$= \lim_{x \to 1}\frac{(x+1)(x-1)}{(x-1)(\sqrt{24}+\sqrt{25-x^2})}$$ $$= \lim_{x \to 1}\frac{x+1}{\sqrt{24}+\sqrt{25-x^2}} \quad [\because x \to 1 \implies x \neq 1]$$ $$= \frac{1+1}{\sqrt{24}+\sqrt{25-1^2}} = \frac{2}{2\sqrt{24}} = \frac{1}{2\sqrt{6}} \quad \text{(Ans.)}$$

9. Limits as Variables Tend to Infinity

$9(i)$ Evaluate: $$\lim_{x \to \infty} \frac{6-5x^2}{8x-15x^2}$$

Solution

Substituting $x = \frac{1}{h}$. As $x \to \infty$, $h \to 0$:

$$\lim_{x \to \infty} \frac{6-5x^2}{8x-15x^2} = \lim_{h \to 0}\left(\frac{6 - 5/h^2}{8/h - 15/h^2}\right)$$ $$= \lim_{h \to 0}\left(\frac{6h^2-5}{8h-15}\right) = \frac{6(0)-5}{8(0)-15} = \frac{-5}{-15} = \frac{1}{3} \quad \text{(Ans.)}$$

$9(ii)$ Evaluate: $$\lim_{x \to \infty}\frac{2x-5}{3x+2}$$

Solution

Substituting $x = \frac{1}{h}$. As $x \to \infty$, $h \to 0$:

$$\lim_{x \to \infty}\frac{2x-5}{3x+2} = \lim_{h \to 0}\left(\frac{2/h-5}{3/h+2}\right)$$ $$= \lim_{h \to 0}\left(\frac{2-5h}{3+2h}\right) = \frac{2-5(0)}{3+2(0)} = \frac{2}{3} \quad \text{(Ans.)}$$

$9(iii)$ Evaluate: $$\lim_{n \to \infty} \frac{3n^2-4n+6}{n^2+6n-7}$$

Solution

Substituting $n = \frac{1}{z}$. As $n \to \infty$, $z \to 0$:

$$\lim_{n \to \infty} \frac{3n^2-4n+6}{n^2+6n-7} = \lim_{z \to 0}\left(\frac{3/z^2-4/z+6}{1/z^2+6/z-7}\right)$$ $$= \lim_{z \to 0}\left(\frac{3-4z+6z^2}{1+6z-7z^2}\right) = \frac{3-0+0}{1+0-0} = 3 \quad \text{(Ans.)}$$

$9(iv)$ Evaluate: $$\lim_{n \to \infty} \frac{1+\sqrt{n}}{1-\sqrt{n}}$$

Solution

Substituting $n = \frac{1}{h}$. As $n \to \infty$, $h \to 0$:

$$\lim_{n \to \infty} \frac{1+\sqrt{n}}{1-\sqrt{n}} = \lim_{h \to 0}\left(\frac{1+1/\sqrt{h}}{1-1/\sqrt{h}}\right)$$ $$= \lim_{h \to 0}\left(\frac{\sqrt{h}+1}{\sqrt{h}-1}\right) = \frac{\sqrt{0}+1}{\sqrt{0}-1} = -1 \quad \text{(Ans.)}$$

$9(v)$ Evaluate: $$\lim_{x \to \infty}\frac{4x^3-3x^2+6x-2}{3+5x^2-5x^3}$$

Solution

Substituting $x = \frac{1}{y}$. As $x \to \infty$, $y \to 0$:

$$\lim_{x \to \infty}\frac{4x^3-3x^2+6x-2}{3+5x^2-5x^3} = \lim_{y \to 0}\left(\frac{4/y^3-3/y^2+6/y-2}{3+5/y^2-5/y^3}\right)$$ $$= \lim_{y \to 0}\left(\frac{4-3y+6y^2-2y^3}{3y^3+5y-5}\right) = \frac{4-0+0-0}{0+0-5} = -\frac{4}{5} \quad \text{(Ans.)}$$

$9(vi)$ Evaluate: $$\lim_{x \to \infty} e^{\frac{1}{x^2}+2}$$

Solution

Substituting $x = \frac{1}{y}$. As $x \to \infty$, $y \to 0$:

$$\lim_{x \to \infty} e^{\frac{1}{x^2}+2} = \lim_{y \to 0} e^{y^2+2} = e^{0^2+2} = e^2 \quad \text{(Ans.)}$$

$9(vii)$ Evaluate: $$\lim_{n \to \infty} \frac{2n^2+3n-9}{3n^3+2n+7}$$

Solution

Substituting $n = \frac{1}{y}$. As $n \to \infty$, $y \to 0$:

$$\lim_{n \to \infty} \frac{2n^2+3n-9}{3n^3+2n+7} = \lim_{y \to 0}\frac{2/y^2+3/y-9}{3/y^3+2/y+7}$$

Multiplying numerator and denominator by $y^3$:

$$= \lim_{y \to 0}\frac{2y+3y^2-9y^3}{3+2y^2+7y^3} = \frac{2(0)+3(0)^2-9(0)^3}{3+2(0)^2+7(0)^3} = \frac{0}{3} = 0 \quad \text{(Ans.)}$$

$9(viii)$ Evaluate: $$\lim_{x \to \infty}\frac{x^4+3x^3-2x^2+5}{x^3-3x^2+2x-1}$$

Solution

$$\lim_{x \to \infty}\frac{x^4+3x^3-2x^2+5}{x^3-3x^2+2x-1} = \lim_{x \to \infty} x \cdot \left(\frac{x^3+3x^2-2x+5/x}{x^3-3x^2+2x-1}\right)$$

Substituting $x = \frac{1}{h}$. As $x \to \infty$, $h \to 0$:

$$= \left(\lim_{x \to \infty} x\right) \times \lim_{h \to 0}\frac{1/h^3+3/h^2-2/h+5h}{1/h^3-3/h^2+2/h-1}$$ $$= \left(\lim_{x \to \infty} x\right) \times \lim_{h \to 0}\frac{1+3h-2h^2+5h^4}{1-3h+2h^2-h^3}$$ $$= \left(\lim_{x \to \infty} x\right) \times \frac{1+0-0+0}{1-0+0-0} = \lim_{x \to \infty} x = \infty \quad \text{(Ans.)}$$

Simplification Note: Numerator and denominator inside the bracket are multiplied by $h^3$ to clear fraction denominators.

$9(ix)$ Evaluate: $$\lim_{n \to \infty}\frac{\sqrt{1+n^2}-\sqrt{1+n}}{\sqrt{1+n^3}-\sqrt{1+n}}$$

Solution

Rationalizing numerator and denominator as in Question $6$:

$$\lim_{n \to \infty}\frac{\sqrt{1+n^2}-\sqrt{1+n}}{\sqrt{1+n^3}-\sqrt{1+n}} = \lim_{n \to \infty}\frac{(\sqrt{1+n^3}+\sqrt{1+n})}{(n+1)(\sqrt{1+n^2}+\sqrt{1+n})}$$

Substituting $n = \frac{1}{h}$. As $n \to \infty$, $h \to 0$:

$$= \lim_{h \to 0}\frac{\sqrt{1+\frac{1}{h^3}}+\sqrt{1+\frac{1}{h}}}{\left(1+\frac{1}{h}\right)\left[\sqrt{1+\frac{1}{h^2}}+\sqrt{1+\frac{1}{h}}\right]}$$ $$= \lim_{h \to 0}\frac{\frac{\sqrt{h^3+1}}{h^{3/2}}+\frac{\sqrt{h+1}}{h^{1/2}}}{\left(\frac{1+h}{h}\right)\left[\frac{\sqrt{h^2+1}}{h}+\frac{\sqrt{h+1}}{h^{1/2}}\right]}$$ $$= \lim_{h \to 0}\frac{\sqrt{h}\sqrt{h^3+1}+h^{3/2}\sqrt{h+1}}{(1+h)(\sqrt{h^2+1}+\sqrt{h}\sqrt{h+1})}$$ $$= \frac{0}{1 \times (1+0)} = 0 \quad \text{(Ans.)}$$