Differentiation (Part-38) | S N De

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

Sol. $\,\lim_{x \to 0} \frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}\\=\lim_{x \to 0}\frac{(\sqrt{1+x^2}-\sqrt{1+x})(\sqrt{1+x^2}+\sqrt{1+x})\\~~~~~~\times (\sqrt{1+x^3}+\sqrt{1+x})}{(\sqrt{1+x^3}-\sqrt{1+x})(\sqrt{1+x^3}+\sqrt{1+x})\\~~~~~ \times(\sqrt{1+x^2}+\sqrt{1+x})}\\=\lim_{x \to 0}\frac{\{(\sqrt{1+x^2})^2-(\sqrt{1+x})^2\}(\sqrt{1+x^3}+\sqrt{1+x})}{\{(\sqrt{1+x^3})^2-(\sqrt{1+x})^2\}(\sqrt{1+x^2}+\sqrt{1+x})}\\=\lim_{x \to 0}\frac{(1+x^2-1-x)(\sqrt{1+x^3}+\sqrt{1+x})}{(1+x^3-1-x)(\sqrt{1+x^2}+\sqrt{1+x})}\\=\lim_{x \to 0}\frac{x(x-1)(\sqrt{1+x^3}+\sqrt{1+x})}{x(x^2-1)(\sqrt{1+x^2}+\sqrt{1+x})}\\=\lim_{x \to 0}\frac{x(x-1)(\sqrt{1+x^3}+\sqrt{1+x})}{x(x+1)(x-1)(\sqrt{1+x^2}+\sqrt{1+x})}\\=\lim_{x \to 0}\frac{(\sqrt{1+x^3}+\sqrt{1+x})}{(x+1)(\sqrt{1+x^2}+\sqrt{1+x})}\\=\frac{\sqrt{1+0^3}+\sqrt{1+0}}{(0+1)(\sqrt{1+0^2}+\sqrt{1+0})}\\=1~~~~\text{(proved)}$

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

Sol.  $\, ~~~\lim_{x \to a}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}\\=\lim_{x \to a}\frac{(\sqrt{a+2x}-\sqrt{3x})(\sqrt{a+2x}+\sqrt{3x})(\sqrt{3a+x}+2\sqrt{x})}{(\sqrt{3a+x}-2\sqrt{x})(\sqrt{3a+x}+2\sqrt{x})(\sqrt{a+2x}+\sqrt{3x})}\\=\lim_{x \to a}\frac{\{(\sqrt{a+2x})^2-(\sqrt{3x})^2\}(\sqrt{3a+x}+2\sqrt{x})}{\{(\sqrt{3a+x})^2-(2\sqrt{x})^2\}(\sqrt{a+2x}+\sqrt{3x})}\\=\lim_{x \to a}\frac{(a+2x-3x)(\sqrt{3a+x}+2\sqrt x)}{(3a+x-4x)(\sqrt{a+2x}+\sqrt{3x})}\\=\lim_{x \to a}\frac{(a-x)(\sqrt{3a+x}+2\sqrt{x})}{3(a-x)(\sqrt{a+2x}+\sqrt{3x})}\\=\lim_{x \to a}\frac{(\sqrt{3a+x}+2\sqrt{x})}{3(\sqrt{a+2x}+\sqrt{3x})}\\=\frac{\sqrt{3a+a}+2\sqrt a}{3(\sqrt{a+2a}+\sqrt{3a})}\\=\frac{\sqrt{4a}+2\sqrt a}{3(\sqrt{3a}+\sqrt{3a})}\\=\frac{2\sqrt a+2\sqrt a}{3 \times 2\sqrt{3a}}\\=\frac{4\sqrt a}{6\sqrt{3a}}\\=\frac{2}{3\sqrt 3}\,\,\text{(ans.)}$

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

Sol. $\,~~~\lim_{x \to 1}\frac{G(x)-G(1)}{x-1}\\=\lim_{x \to 1}\frac{-\sqrt{25-x^2}~~+\sqrt{25-1^2}}{x-1}\\=\lim_{x \to 1}\frac{(\sqrt{24}-\sqrt{25-x^2})(\sqrt{24}+\sqrt{25-x^2})}{(x-1)(\sqrt{24}+\sqrt{25-x^2})}\\=\lim_{x \to 1}\frac{(\sqrt{24})^2-(\sqrt{25-x^2})^2}{(x-1)(\sqrt{24}+\sqrt{25-x^2})}\\=\lim_{x \to 1}\frac{24-25+x^2}{(x-1)(\sqrt{24}+\sqrt{25-x^2})}\\=\lim_{x \to 1}\frac{x^2-1}{(x-1)(\sqrt{24}+\sqrt{25-x^2})}\\=\lim_{x \to 1}\frac{(x+1)(x-1)}{(x-1)(\sqrt{24}+\sqrt{25-x^2})}\\=\lim_{x \to 1}\frac{x+1}{(\sqrt{24}+\sqrt{25-x^2})}~~[\because x \to 1, x \neq 1]\\=\frac{1+1}{\sqrt{24}+\sqrt{25-1^2}}\\=\frac{2}{2\sqrt{4 \times 6}}\\=\frac{2}{2 \times 2\sqrt{6}}\\=\frac{1}{2\sqrt 6}\,\,\,\text{(ans.)}$

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

Sol. $\,\,\lim_{x \to \infty} \frac{6-5x^2}{8x-15x^2}\\=\lim_{h \to 0}\left(\frac{6-5/h^2}{8/h-15/h^2}\right)\\~~[\text{let}~~x=\frac 1h,\,\text{as}~~x \to \infty, h \to 0]\\=\lim_{h \to 0} \left(\frac{6h^2-5}{8h-15}\right)\\=\frac{6 \cdot 0^2-5}{8 \cdot 0-15}\\=\frac{-5}{-15}\\=\frac{1}{3}\,\,\text{(ans.)}$

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

Sol. $\,\,~~\lim_{x \to \infty}\frac{2x-5}{3x+2}\\=\lim_{h \to 0}\left(\frac{2/h-5}{3/h+2}\right)\\~~[\text{let}~~x=\frac 1h,\,\text{as}~~x \to \infty, h \to 0]\\=\lim_{h \to 0}\left(\frac{2-5h}{3+2h}\right)\\=\frac{2-5 \cdot 0}{3+2 \cdot 0}\\=\frac 23\,\,\text{(ans.)}$

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

Sol. $\,\,\lim_{n \to \infty} \frac{3n^2-4n+6}{n^2+6n-7}\\=\lim_{z \to 0}\left(\frac{3/z^2-4/z+6}{1/z^2+6/z-7}\right)\\~~[\text{let}~~n=\frac 1z,\,\text{as}~~n \to \infty, z \to 0]\\=\lim_{z \to 0}\left(\frac{3-4z+6z^2}{1+6z-7z^2}\right)\\=\frac{3-4 \cdot 0+6 \cdot 0^2}{1+6\cdot 0-7 \cdot 0^2}\\=\frac 31\\=3\,\,\text{(ans.)}$

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

Sol. $\,\,\lim_{n \to \infty} \frac{1+\sqrt n}{1-\sqrt n}\\=\lim_{h \to 0}\left(\frac{1+\frac{1}{\sqrt h}}{1-\frac{1}{\sqrt h}}\right)\\~~[\text{let}~~n=\frac 1h,\,\text{as}~~ \to \infty, h \to 0]\\=\lim_{h \to 0}\left(\frac{\sqrt h+1}{\sqrt h-1}\right)\\=\frac{\sqrt 0+1}{\sqrt 0-1}\\=\frac{1}{-1}\\=-1\,\,\text{(ans.)}$

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

Sol. $\,\,\lim_{x \to \infty}\frac{4x^3-3x^2+6x-2}{3+5x^2-5x^3}\\=\lim_{y \to 0}\frac{4/y^3-3/y^2+6/y-2}{3+5/y^2-5/y^3}\\~~[\text{let}~~x=\frac 1y,\,\text{as}~~x \to \infty, y \to 0]\\=\lim_{y \to 0}\frac{4-3y+6y^2-2y^3}{3y^3+5y-5}\\=\frac{4-3 \times 0+6\times 0^2-2 \times 0^3}{3 \times 0^3+5 \times 0-5}\\=\frac{4}{-5}\\=-\frac 45\,\,\,\text{(ans.)}$

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

Sol. $\,\,\lim_{x \to \infty}~~ e^{\frac{1}{x^2}+2}\\=\lim_{y \to 0}~~e^{y^2+2}\\~~[\text{let}~~x=\frac 1y,\,\text{as}~~x \to \infty, y \to 0]\\=e^{0^2+2}\\=e^2\,\,\text{(ans.)}$

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

Sol. $\,\,\lim_{n \to \infty} \frac{2n^2+3n-9}{3n^3+2n+7}\\=\lim_{y \to 0}\frac{2/y^2+3/y-9}{3/y^3+2/y+7}\\~~[\text{let}~~n=\frac 1y,\,\text{as}~~n \to \infty, y \to 0]\\=\lim_{y \to 0}\frac{y^3(2/y^2+3/y-9)}{y^3(3/y^3+2/y+7)}\\=\lim_{y \to 0}\frac{2y+3y^2-9y^3}{3+2y^2+7y^3}\\=\frac{2 \cdot 0+3 \cdot 0^2-9 \cdot 0^3}{3+2 \cdot 0^2+7 \cdot 0^3}\\=\frac 03\\=0\,\,\text{(ans.)}$

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

Sol. $\,\,\lim_{x \to \infty}\frac{x^4+3x^3-2x^2+5}{x^3-3x^2+2x-1}\\=\lim_{x \to \infty}\frac{x(x^3+3x^2-2x+5/x)}{x^3-3x^2+2x-1}\\=\lim_{x \to \infty}x \times \lim_{h \to 0}\frac{1/h^3+3/h^2-22/h+5h}{1/h^3-3/h^2+2/h-1}\\~~[\text{let}~~x=\frac 1h,\,\text{as}~~x \to \infty, h \to 0]\\=\lim_{x \to \infty}x \times \lim_{h \to 0}\frac{1+3h-22h^2+5h^4}{1-3h+2h^2-h^3}~~[*]\\=(\lim_{x \to \infty} x )\times \frac{1+3 \cdot 0 -22 \cdot  0^2+5 \cdot  0^4}{1-3 \cdot  0+2 \cdot  0^2-0^3}\\= (\lim_{x \to \infty} x)  \times 1\\=\lim_{x \to \infty} x \\=\infty\,\,\text{(ans.)}$

Note[*]: Multiplying numerator and denominator by $\,h^3.$

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

Sol. $\,\,\lim_{n \to \infty}\frac{\sqrt{1+n^2}-\sqrt{1+n}}{\sqrt{1+n^3}-\sqrt{1+n}}\\=\lim_{n \to \infty}\frac{(\sqrt{1+n^2}-\sqrt{1+n})(\sqrt{1+n^2}+\sqrt{1+n})\\~~~\times (\sqrt{1+n^3}+\sqrt{1+n})}{(\sqrt{1+n^3}-\sqrt{1+n})(\sqrt{1+n^3}+\sqrt{1+n})\\~~~\times(\sqrt{1+n^2}+\sqrt{1+n})}\\=\lim_{n \to \infty}\frac{\{(\sqrt{1+n^2})^2-(\sqrt{1+n})^2\}(\sqrt{1+n^3}+\sqrt{1+n})}{\{(\sqrt{1+n^3})^2-(\sqrt{1+n})^2\}(\sqrt{1+n^2}+\sqrt{1+n})}\\=\lim_{n \to \infty}\frac{(1+n^2-1-n)(\sqrt{1+n^3}+\sqrt{1+n})}{(1+n^3-1-n)(\sqrt{1+n^2}+\sqrt{1+n})}\\=\lim_{n \to \infty}\frac{n(n-1)(\sqrt{1+n^3}+\sqrt{1+n})}{n(n+1)(n-1)(\sqrt{1+n^2}+\sqrt{1+n})}\\=\lim_{n \to \infty}\frac{(\sqrt{1+n^3}+\sqrt{1+n})}{(n+1)(\sqrt{1+n^2}+\sqrt{1+n})}\\=\lim_{h  \to 0}\frac{\sqrt{1+\frac{1}{h^3}}+\sqrt{1+\frac 1h}}{\left(1+\frac 1h\right)\left[\sqrt{1+\frac{1}{h^2}}+\sqrt{1+\frac 1h}\right]}\\ ~~[\text{let}~~n=\frac 1h,\,\text{as}~~n \to \infty, h \to 0]\\=\lim_{h  \to 0}\frac{\frac{\sqrt{h^3+1}}{h^{3/2}}+\frac{\sqrt{h+1}}{h^{1/2}}}{\left(\frac{1+h}{h}\right)\left[\frac{\sqrt{h^2+1}}{h}+\frac{\sqrt{h+1}}{h^{1/2}}\right]}\\=\lim_{h  \to 0}\frac{\sqrt h \sqrt{h^3+1}+h^{3/2}\sqrt{h+1}}{(1+h)(\sqrt{h^2+1}+\sqrt h\sqrt{h+1})}\\=\frac{0}{1 \times (1+0)}\\=0\,\,\text{(ans.)}$