Question

lim x tends to 3 √2x+3 -√4x-3/ x^2 -9​

Answer 1

We're asked to evaluate,

$\displaystyle\longrightarrow L=\lim_{x\to3}\dfrac{\sqrt{2x+3}-\sqrt{4x-3}}{x^2-9}$

On taking $x=3$ directly we get indeterminate form.

$\displaystyle\longrightarrow\dfrac{\sqrt{2\cdot3+3}-\sqrt{4\cdot3-3}}{3^2-9}=\dfrac{0}{0}$

Then by L'Hospital's Rule,

$\displaystyle\longrightarrow L=\lim_{x\to3}\dfrac{\dfrac{2}{2\sqrt{2x+3}}-\dfrac{4}{2\sqrt{4x-3}}}{2x}$

$\displaystyle\longrightarrow L=\lim_{x\to3}\dfrac{\dfrac{1}{\sqrt{2x+3}}-\dfrac{2}{\sqrt{4x-3}}}{2x}$

$\displaystyle\longrightarrow L=\lim_{x\to3}\dfrac{\sqrt{4x-3}-2\sqrt{2x+3}}{2x\sqrt{(2x+3)(4x-3)}}$

Taking $x=3,$

$\displaystyle\longrightarrow L=\dfrac{\sqrt{4\cdot3-3}-2\sqrt{2\cdot3+3}}{2\cdot3\sqrt{(2\cdot3+3)(4\cdot3-3)}}$

$\displaystyle\longrightarrow\underline{\underline{L=-\dfrac{1}{18}}}$