Differentiation (Part-38) | S N De

 

$\,11.\,$ Evaluate the following limits :

$\,(i)\,\,\lim_{x \to 0}~~\frac{\sqrt{1+x}-\sqrt{1-x}}{x}$

Sol. $\,\,\lim_{x \to 0}~~\frac{\sqrt{1+x}-\sqrt{1-x}}{x}\\=\lim_{x \to 0}\frac{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x}+\sqrt{1-x})}{x(\sqrt{1+x}+\sqrt{1-x})}\\=\lim_{x \to 0}\frac{(\sqrt{1+x})^2-(\sqrt{1-x})^2}{x(\sqrt{1+x}+\sqrt{1-x})}\\=\lim_{x \to 0}\frac{1+x-1+x}{x(\sqrt{1+x}+\sqrt{1-x})}\\=\lim_{x \to 0}\frac{2x}{x(\sqrt{1+x}+\sqrt{1-x})}\\=\lim_{x \to 0}\frac{2}{(\sqrt{1+x}+\sqrt{1-x})}\\=\frac{2}{(\sqrt{1+0}+\sqrt{1-0})}\\=\frac 22 \\=1~~\text{(ans.)}$

$\,(ii)\,\lim_{x  \to 0}\frac{1-\sqrt{1-x^2}}{x^2}$

Sol. $\,\lim_{x  \to 0}\,\frac{1-\sqrt{1-x^2}}{x^2}\\=\lim_{x  \to 0}\frac{(1-\sqrt{1-x^2})(1+\sqrt{1-x^2})}{x^2(1+\sqrt{1-x^2})}\\=\lim_{x  \to 0}\frac{1^2-(\sqrt{1-x^2})^2}{x^2(1+\sqrt{1-x^2})}\\=\lim_{x  \to 0}\frac{1-1+x^2}{x^2(1+\sqrt{1-x^2})}\\=\lim_{x  \to 0}\frac{1}{(1+\sqrt{1-x^2})}\\=\frac{1}{(1+\sqrt{1-0^2})}\\=\frac 12\,\,\text{(ans.)}$

$\,(iii)\lim_{h \to 1}\,\frac{h-1}{\sqrt h-1}$

Sol. $~~~\,\lim_{h \to 1}\frac{h-1}{\sqrt h-1}\\=\lim_{h \to 1}\frac{(\sqrt h)^2-1^2}{(\sqrt h-1)}\\=\lim_{h \to 1}\frac{(\sqrt h+1)(\sqrt h-1)}{(\sqrt h-1)}\\=\lim_{h \to 1}\frac{\sqrt h+1}{1}\\=\sqrt 1+1\\=2\,\,\text{(ans.)}$

$\,(iv)\,\lim_{x  \to 2}\, \frac{\sqrt x-\sqrt 2}{x^2-4}$

Sol. $~~\lim_{x  \to 2}\, \frac{\sqrt x-\sqrt 2}{x^2-4}\\=\lim_{x  \to 2}\frac{(\sqrt x -\sqrt 2)}{(x)^2-(2)^2}\\=\lim_{x  \to 2}\frac{(\sqrt x -\sqrt 2)}{(x+2)(x-2)}\\=\lim_{x  \to 2}\frac{(\sqrt x -\sqrt 2)}{(x+2)(\sqrt x +\sqrt 2)(\sqrt x -\sqrt 2)}\\=\lim_{x  \to 2}\frac{1}{(x+2)(\sqrt x +\sqrt 2)}\\=\frac{1}{(2+2)(\sqrt 2+\sqrt 2)}\\=\frac{1}{4 \times 2\sqrt 2}\\=\frac{1}{8\sqrt 2}~~~\text{(ans.)}$

Note : $\, x-2=(x)^2-(\sqrt 2)^2=(x+\sqrt 2)(x-\sqrt 2)$

$\,(v)\,\lim_{x  \to 0}\, \frac{x^2}{a-\sqrt{a^2-x^2}}$

Sol. $~~\lim_{x  \to 0}\, \frac{x^2}{a-\sqrt{a^2-x^2}}\\=\lim_{x  \to 0}\frac{x^2(a+\sqrt{a^2-x^2})}{(a-\sqrt{a^2-x^2})(a+\sqrt{a^2-x^2})}\\=\lim_{x  \to 0} \frac{x^2(a+\sqrt{a^2-x^2})}{a^2-(\sqrt{a^2-x^2})^2}\\=\lim_{x  \to 0}\frac{x^2(a+\sqrt{a^2-x^2})}{a^2-a^2+x^2}\\=\lim_{x  \to 0} \frac{x^2(a+\sqrt{a^2-x^2})}{x^2}\\=\lim_{x  \to 0}(a+\sqrt{a^2-x^2})\\=a+\sqrt{a^2-0^2}\\=a+a\\=2a~~~\text{(ans.)}$

$\,(vi)\,\lim_{x \to 0}\,\frac{\sqrt{a+x}-\sqrt a}{x}$

Sol. $\,\,\lim_{x \to 0}\,\frac{\sqrt{a+x}-\sqrt a}{x}\\=\lim_{x \to 0}\frac{(\sqrt{a+x}+\sqrt a)(\sqrt{a+x}-\sqrt a)}{x(\sqrt{a+x}+\sqrt a)}\\=\lim_{x \to 0}\frac{(\sqrt{a+x})^2-(\sqrt{a})^2}{x(\sqrt{a+x}+\sqrt a)}\\=\lim_{x \to 0}\frac{a+x-a}{x(\sqrt{a+x}+\sqrt a)}\\=\lim_{x \to 0}\frac{x}{x(\sqrt{a+x}+\sqrt a)}\\=\lim_{x \to 0}\frac{1}{\sqrt{a+x}+\sqrt a}\\=\frac{1}{\sqrt{a+0}+\sqrt a}\\=\frac{1}{2\sqrt a}\,\,\text{(ans.)}$

$\,(vii)\lim_{x \to 3}\,\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}$

Sol. $\,\lim_{x \to 3}\,\frac{x-3}{\sqrt{x-2}-\sqrt{4-x}}\\=\lim_{x \to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{(\sqrt{x-2}-\sqrt{4-x})(\sqrt{x-2}+\sqrt{4-x})}\\=\lim_{x \to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{(\sqrt{x-2})^2-(\sqrt{4-x})^2}\\=\lim_{x \to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{x-2-4+x}\\=\lim_{x \to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{2x-6}\\=\lim_{x \to 3}\frac{(x-3)(\sqrt{x-2}+\sqrt{4-x})}{2(x-3)}\\=\lim_{x \to 3}~~\frac 12(\sqrt{x-2}+\sqrt{4-x})\\=\frac 12 (\sqrt{3-2}+\sqrt{4-3})\\=\frac 12(1+1)\\=1\,\,\text{(ans.)}$

$\,(viii)\,~~~\lim_{x \to a}\frac{\sqrt{x-b}-\sqrt{a-b}}{x^2-a^2}$

$\,~~~\lim_{x \to a}\frac{\sqrt{x-b}-\sqrt{a-b}}{x^2-a^2}\\=\lim_{x \to a}\frac{(\sqrt{x-b}+\sqrt{a-b})(\sqrt{x-b}-\sqrt{a-b})}{(x+a)(x-a)(\sqrt{x-b}+\sqrt{a-b})}\\=\lim_{x \to a}\frac{(\sqrt{x-b})^2-(\sqrt{a-b})^2}{(x+a)(x-a)(\sqrt{x-b}+\sqrt{a-b})}\\=\lim_{x \to a}\frac{x-b-a+b}{(x+a)(x-a)(\sqrt{x-b}+\sqrt{a-b})}\\=\lim_{x \to a}\frac{(x-a)}{(x+a)(x-a)(\sqrt{x-b}+\sqrt{a-b})}\\=\lim_{x \to a}\frac{1}{(x+a)(\sqrt{x-b}+\sqrt{a-b})}\\=\frac{1}{(a+a)(\sqrt{a-b}+\sqrt{a-b})}\\=\frac{1}{2a \times 2\sqrt{a-b}}\\=\frac{1}{4a\sqrt{a-b}}\,\,\text{(ans.)}$

$\,(ix)\lim_{x  \to 1}\frac{x^2-1}{\sqrt{3x+1}-\sqrt{5x-1}}$

Sol. $\,~~~\lim_{x  \to 1}\frac{x^2-1}{\sqrt{3x+1}-\sqrt{5x-1}}\\=\lim_{x  \to 1}\frac{(x+1)(x-1)(\sqrt{3x+1}+\sqrt{5x-1})}{(\sqrt{3x+1}+\sqrt{5x-1})(\sqrt{3x+1}-\sqrt{5x-1})}\\=\lim_{x  \to 1}\frac{(x+1)(x-1)(\sqrt{3x+1}+\sqrt{5x-1})}{(\sqrt{3x+1})^2-(\sqrt{5x-1})^2}\\=\lim_{x  \to 1}\frac{(x+1)(x-1)(\sqrt{3x+1}+\sqrt{5x-1})}{3x+1-5x+2}\\=\lim_{x  \to 1}\frac{(x+1)(x-1)(\sqrt{3x+1}+\sqrt{5x-1})}{-2(x-1)}\\=-\frac 12 \times \lim_{x  \to 1} [(x+1)(\sqrt{3x+1}+\sqrt{5x-1})]\\=-\frac 12 \times [(1+1)(\sqrt{3 \times 1 +1}+\sqrt{5 \times 1-1})]\\=-\frac 12 \times [2 \times (\sqrt 4+\sqrt 4)]\\=-\frac 12 \times [2 \times (2+2)]\\=-4\,\,\text{(ans.)}$

$\,~(x)\lim_{x  \to 2}\frac{\sqrt{1+2x}-\sqrt{1+x^2}}{2-x}$

Sol. $\,~~~~\lim_{x  \to 2}\frac{\sqrt{1+2x}-\sqrt{1+x^2}}{2-x}\\=\lim_{x  \to 2}\frac{(\sqrt{1+2x}-\sqrt{1+x^2})(\sqrt{1+2x}+\sqrt{1+x^2})}{(2-x)(\sqrt{1+2x}+\sqrt{1+x^2})}\\=\lim_{x  \to 2}\frac{(\sqrt{1+2x})^2-(\sqrt{1+x^2})^2}{(2-x)(\sqrt{1+2x}+\sqrt{1+x^2})}\\=\lim_{x  \to 2}\frac{1+2x-1-x^2}{(2-x)(\sqrt{1+2x}+\sqrt{1+x^2})}\\=\lim_{x  \to 2}\frac{x(2-x)}{(2-x)(\sqrt{1+2x}+\sqrt{1+x^2})}\\=\lim_{x  \to 2}\frac{x}{\sqrt{1+2x}+\sqrt{1+x^2}}\\~~~~~[\because x \to 2,\,\, (2-x) \neq 0]\\=\frac{2}{\sqrt{1+2 \times 2}+\sqrt{1+2^2}}\\=\frac{2}{\sqrt{5}+\sqrt 5}\\=\frac{2}{2\sqrt 5}\\=\frac{1}{\sqrt 5}\,\,\text{(ans.)}$

$\,(xi)~\lim_{x  \to 4}\frac{\sqrt x-2}{x-4}$

Sol. $~~~\,\lim_{x  \to 4}\frac{\sqrt x-2}{x-4}\\=\lim_{x  \to 4}\frac{\sqrt x-2}{(\sqrt x)^2-2^2}\\=\lim_{x  \to 4}\frac{(\sqrt x-2)}{(\sqrt x+2)(\sqrt x-2)}\\=\lim_{x  \to 4}\frac{1}{\sqrt x+2}\\=\frac{1}{\sqrt 4+2}\\=\frac{1}{2+2}\\=\frac 14\,\,\text{(ans.)}$

 $(xii)~~\,\lim_{x  \to 0}\frac{\sqrt{1+x+x^2}-1}{x}$

Sol. $\,~~~\lim_{x  \to 0}\frac{\sqrt{1+x+x^2}-1}{x}\\=\lim_{x  \to 0}\frac{(\sqrt{1+x+x^2}-1)(\sqrt{1+x+x^2}+1)}{x(\sqrt{1+x+x^2}+1)}\\=\lim_{x  \to 0}\frac{(\sqrt{1+x+x^2})^2-1}{x(\sqrt{1+x+x^2}+1)}\\=\lim_{x  \to 0}\frac{1+x+x^2-1}{x(\sqrt{1+x+x^2}+1)}\\=\lim_{x  \to 0}\frac{x(1+x)}{x(\sqrt{1+x+x^2}+1)}\\=\lim_{x  \to 0}\frac{1+x}{\sqrt{1+x+x^2}+1}\\=\frac{1+0}{\sqrt{1+0+0^2}+1}\\=\frac 12\,\,\text{(ans.)}$

$\,(xiii)~~\lim_{x  \to 0}\frac{\sqrt{1+2x^2}-\sqrt{1-2x^2}}{x^2}$

Sol. $\,~~\lim_{x  \to 0}~~\frac{\sqrt{1+2x^2}-\sqrt{1-2x^2}}{x^2}\\=\lim_{x  \to 0}\frac{(\sqrt{1+2x^2}+\sqrt{1-2x^2})(\sqrt{1+2x^2}-\sqrt{1-2x^2})}{x^2(\sqrt{1+2x^2}+\sqrt{1-2x^2})}\\=\lim_{x  \to 0}\frac{(\sqrt{1+2x^2})^2-(\sqrt{1-2x^2})^2}{x^2(\sqrt{1+2x^2}+\sqrt{1-2x^2})}\\=\lim_{x  \to 0}\frac{1+2x^2-1+2x^2}{x^2(\sqrt{1+2x^2}+\sqrt{1-2x^2})}\\=\lim_{x  \to 0}\frac{4x^2}{x^2(\sqrt{1+2x^2}+\sqrt{1-2x^2})}\\=\frac{4}{\sqrt{1+2\times 0^2}+\sqrt{1-2 \times 0^2}}\\=\frac{4}{1+1}\\=\frac 42\\=2\,\,\text{(ans.)}$