Question

Lim x tends to 3 x-3/√(x-2)-√ (4-x)

Answer 1

Given,

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

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

$\displaystyle\longrightarrow L=\dfrac{3-3}{\sqrt{3-2}-\sqrt{4-3}}$

$\displaystyle\longrightarrow L=\dfrac{0}{0}$

Then by L'hospital's Rule,

$\displaystyle\longrightarrow L=\lim_{x\to3}\dfrac{1}{\left(\dfrac{1}{2\sqrt{x-2}}\cdot1-\dfrac{1}{2\sqrt{4-x}}\cdot-1\right)}$

$\displaystyle\longrightarrow L=2\lim_{x\to3}\dfrac{1}{\left(\dfrac{1}{\sqrt{x-2}}+\dfrac{1}{\sqrt{4-x}}\right)}$

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

Taking $x=3,$

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

$\displaystyle\longrightarrow L=2\cdot\dfrac{1}{2}$

$\displaystyle\longrightarrow\underline{\underline{L=1}}$

Hence 1 is the answer.