Question

Lim x ➡️ 3 , √x+1 -2 / √x+6 -3 ,what is the value of x ?

Answer 1

$\textsf{\large{\underline{Solution}:}}$

We have to evaluate:

$= \displaystyle \rm \lim_{x \to3} \bigg( \dfrac{ \sqrt{x + 1} - 2}{ \sqrt{x + 6} - 3 } \bigg)$

On evaluating this limit, we will get an indeterminate form. So, we have to apply L'Hospital's rule.

Applying L'Hospital's rule, we get:

$= \displaystyle \rm \lim_{x \to3} \bigg( \dfrac{ \dfrac{d}{dx} (\sqrt{x + 1} - 2)}{ \dfrac{d}{dx} (\sqrt{x + 6} - 3) } \bigg)$

$= \displaystyle \rm \lim_{x \to3} \bigg( \dfrac{ \dfrac{1}{2 \sqrt{x +1 } } }{ \dfrac{1}{2 \sqrt{x + 6} } } \bigg)$

$= \displaystyle \rm \lim_{x \to3} \bigg( \dfrac{ \sqrt{x + 6} }{ \sqrt{x + 1} } \bigg)$

Put x = 3, we get:

$= \displaystyle \rm\dfrac{ \sqrt{3+ 6} }{ \sqrt{3 + 1} }$

$= \displaystyle \rm\dfrac{ \sqrt{9} }{ \sqrt{4} }$

$\rm = \dfrac{3}{2}$

Therefore:

$: \longmapsto \displaystyle \rm \lim_{x \to3} \bigg( \dfrac{ \sqrt{x + 1} - 2}{ \sqrt{x + 6} - 3 } \bigg) = \dfrac{3}{2}$

$\textsf{\large{\underline{Learn More}:}}$

$\rm 1. \: \: \dfrac{d}{dx} \sin(x) = \cos(x)$

$\rm 2. \: \: \dfrac{d}{dx} \cos(x) = - \sin(x)$

$\rm 3. \: \: \dfrac{d}{dx} \tan(x) = \sec^{2} (x)$

$\rm 4. \: \: \dfrac{d}{dx} \cot(x) = - \csc^{2} (x)$

$\rm 5. \: \: \dfrac{d}{dx} \sec(x) = \tan(x) \cdot \sec(x)$

$\rm 6. \: \: \dfrac{d}{dx} \csc(x) = - \cot(x) \cdot \csc(x)$

$\rm 7. \: \: \dfrac{d}{dx} {x}^{n} = n {x}^{n - 1}$

$\rm 8. \: \: \dfrac{d}{dx}c=0$

$\rm 9. \: \: \dfrac{d}{dx} {e}^{x} = {e}^{x}$

$\rm 10. \: \: \dfrac{d}{dx} {a}^{x} = {a}^{x} \: ln(a)$

Answer 2

Step-by-step explanation:

$]\textsf{\large{\underline{Solution}:}}$

We have to evaluate:

$= \displaystyle \rm \lim_{x \to3} \bigg( \dfrac{ \sqrt{x + 1} - 2}{ \sqrt{x + 6} - 3 } \bigg)$

On evaluating this limit, we will get an indeterminate form. So, we have to apply L'Hospital's rule.

Applying L'Hospital's rule, we get:

$= \displaystyle \rm \lim_{x \to3} \bigg( \dfrac{ \dfrac{d}{dx} (\sqrt{x + 1} - 2)}{ \dfrac{d}{dx} (\sqrt{x + 6} - 3) } \bigg)$

$= \displaystyle \rm \lim_{x \to3} \bigg( \dfrac{ \dfrac{1}{2 \sqrt{x +1 } } }{ \dfrac{1}{2 \sqrt{x + 6} } } \bigg)$

$= \displaystyle \rm \lim_{x \to3} \bigg( \dfrac{ \sqrt{x + 6} }{ \sqrt{x + 1} } \bigg)$

Put x = 3, we get:

$= \displaystyle \rm\dfrac{ \sqrt{3+ 6} }{ \sqrt{3 + 1} }$

$= \displaystyle \rm\dfrac{ \sqrt{9} }{ \sqrt{4} }$

$\rm = \dfrac{3}{2}$

Therefore:

$: \longmapsto \displaystyle \rm \lim_{x \to3} \bigg( \dfrac{ \sqrt{x + 1} - 2}{ \sqrt{x + 6} - 3 } \bigg) = \dfrac{3}{2}$

$\textsf{\large{\underline{Learn More}:}}$

$\rm 1. \: \: \dfrac{d}{dx} \sin(x) = \cos(x)$

$\rm 2. \: \: \dfrac{d}{dx} \cos(x) = - \sin(x)$

$\rm 3. \: \: \dfrac{d}{dx} \tan(x) = \sec^{2} (x)$

$\rm 4. \: \: \dfrac{d}{dx} \cot(x) = - \csc^{2} (x)$

$\rm 5. \: \: \dfrac{d}{dx} \sec(x) = \tan(x) \cdot \sec(x)$

$\rm 6. \: \: \dfrac{d}{dx} \csc(x) = - \cot(x) \cdot \csc(x)$

$\rm 7. \: \: \dfrac{d}{dx} {x}^{n} = n {x}^{n - 1}$

$\rm 8. \: \: \dfrac{d}{dx}c=0$

$\rm 9. \: \: \dfrac{d}{dx} {e}^{x} = {e}^{x}$

$\rm 10. \: \: \dfrac{d}{dx} {a}^{x} = {a}^{x} \: ln(a)$[/tex]