Question

evaluate the following without using LHospitals rule​

Answer 1

Given,

$\displaystyle\longrightarrow\sf{\lim_{x\to0}\dfrac{\sqrt{a+x}-\sqrt a}{x\sqrt{a^2+ax}}=\ ?}$

On directly taking x as 0 we get indeterminate form.

$\displaystyle\longrightarrow\sf{\dfrac{\sqrt{a+0}-\sqrt a}{0\sqrt{a^2+a\cdot0}}=\dfrac{\sqrt a-\sqrt a}{0\sqrt{a^2}}=\dfrac{0}{0}}$

What L'hospital's Rule says is that,

$\displaystyle\longrightarrow\sf{\lim_{x\to a}\dfrac{f(x)}{g(x)}=\lim_{x\to a}\dfrac{f'(x)}{g'(x)}\ \iff\ \lim_{x\to a}\dfrac{f(x)}{g(x)}=\dfrac{0}{0}\quad OR\quad\lim_{x\to0}\dfrac{f(x)}{g(x)}=\dfrac{\pm\infty}{\pm\infty}}$

Hence,

$\displaystyle\longrightarrow\sf{\lim_{x\to0}\dfrac{\sqrt{a+x}-\sqrt a}{x\sqrt{a^2+ax}}=\lim_{x\to0}\dfrac{\dfrac{d}{dx}\left[\sqrt{a+x}-\sqrt a\right]}{\dfrac{d}{dx}\,\left[x\sqrt{a^2+ax}\right]}}$

$\displaystyle\longrightarrow\sf{\lim_{x\to0}\dfrac{\sqrt{a+x}-\sqrt a}{x\sqrt{a^2+ax}}=\lim_{x\to0}\dfrac{\dfrac{1}{2\sqrt{a+x}}(0+1)-0}{1\sqrt{a^2+ax}+x\left(\dfrac{1}{2\sqrt{a^2+ax}}(0+a)\right)}}$

$\displaystyle\longrightarrow\sf{\lim_{x\to0}\dfrac{\sqrt{a+x}-\sqrt a}{x\sqrt{a^2+ax}}=\lim_{x\to0}\dfrac{\dfrac{1}{2\sqrt{a+x}}}{\sqrt{a^2+ax}+\dfrac{ax}{2\sqrt{a^2+ax}}}}$

$\displaystyle\longrightarrow\sf{\lim_{x\to0}\dfrac{\sqrt{a+x}-\sqrt a}{x\sqrt{a^2+ax}}=\lim_{x\to0}\dfrac{\dfrac{\sqrt a}{2\sqrt{a^2+ax}}}{\left(\dfrac{2(a^2+ax)+ax}{2\sqrt{a^2+ax}}\right)}}$

$\displaystyle\longrightarrow\sf{\lim_{x\to0}\dfrac{\sqrt{a+x}-\sqrt a}{x\sqrt{a^2+ax}}=\lim_{x\to0}\dfrac{\sqrt a}{2a^2+3ax}}$

Taking x as 0 thereby avoiding limit,

$\displaystyle\longrightarrow\sf{\lim_{x\to0}\dfrac{\sqrt{a+x}-\sqrt a}{x\sqrt{a^2+ax}}=\dfrac{\sqrt a}{2a^2+3a(0)}}$

$\displaystyle\longrightarrow\sf{\lim_{x\to0}\dfrac{\sqrt{a+x}-\sqrt a}{x\sqrt{a^2+ax}}=\dfrac{\sqrt a}{2a^2}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{\lim_{x\to0}\dfrac{\sqrt{a+x}-\sqrt a}{x\sqrt{a^2+ax}}=\dfrac{1}{2a\sqrt a}}}}$