Question

Evaluate the limit, $\underset{x\to 3}{\mathop{\lim }}\,\dfrac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}$.
$\begin{align}
  & \text{A) }\dfrac{1}{\sqrt{2}} \\
 & \text{B) }\sqrt{3} \\
 & \text{C) }\dfrac{1}{2\sqrt{2}} \\
 & \text{D) }\dfrac{\sqrt{3}}{\sqrt{2}} \\
\end{align}$

Answer 1

Hint: In this problem we will find the limit by using L’Hospital Rule as the give function whose limit which have to find is in indeterminate form:
Indeterminate form: let f(x) and g(x) be any two function of x such that f(a) = 0 and g(a) = 0, then the ratio $\dfrac{\text{f(x)}}{\text{g(x)}}$ is said to assume the indeterminate form $\dfrac{0}{0}$ at x = a.
There are seven indeterminate form $\dfrac{0}{0},\text{ }\dfrac{\infty }{\infty },\text{ 0}\times \infty ,\text{ }\infty -\infty \text{, }{{\text{0}}^{0}},\text{ }{{\infty }^{0}}\text{ and }{{\text{1}}^{\infty }}.$

Complete step-by-step solution:
L’Hospital Rule: If the function in the indeterminate form then we differentiate the numerator and denominator separately the take limit of numerator and denominator separately if it reduces to indeterminate form then apply the L’Hospital rule again till the limit is obtained.
Let$\text{f(x)}=\dfrac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}$. then
$\Rightarrow \underset{x\to 3}{\mathop{\lim }}\,\text{f(x)}=\underset{x\to 3}{\mathop{\lim }}\,\dfrac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}.....(1)$
Since f(x) has indeterminate form $\dfrac{0}{0}$ at x = 3.
Applying L’hospital rule to equation (1), we get
$\Rightarrow\underset{x\to 3}{\mathop{\lim }}\,\text{f(x)}=\underset{x\to 3}{\mathop{\lim }}\,\dfrac{\dfrac{3}{\sqrt{3x}}}{\dfrac{2}{\sqrt{2x-4}}}$
$\Rightarrow\underset{x\to 3}{\mathop{\lim }}\,\text{f(x)}=\dfrac{\dfrac{3}{\sqrt{3\times 3}}}{\dfrac{2}{\sqrt{2\times 3-4}}}=\dfrac{\dfrac{3}{\sqrt{9}}}{\dfrac{2}{\sqrt{6-4}}}=\dfrac{\dfrac{3}{3}}{\dfrac{2}{\sqrt{2}}}=\dfrac{1}{\sqrt{2}}$
$\Rightarrow\underset{x\to 3}{\mathop{\lim }}\,\text{f(x)}=\dfrac{1}{\sqrt{2}}$
$\Rightarrow\underset{x\to 3}{\mathop{\lim }}\,\dfrac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}=\dfrac{1}{\sqrt{2}}$

Note: In this problem, one should know which function is in indeterminate form and when it is conform that the function is in indeterminate form the use the L’Hospital .
Some of the basic rules of derivative are given below:
If f and g are the two function then
1) Sum rule:
$\dfrac{d}{dx}\left( \text{f+g} \right)=\dfrac{d\text{f}}{dx}+\dfrac{d\text{g}}{dx}$.
2) Product rule:
$\dfrac{d}{dx}\left( \text{f}\cdot \text{g} \right)=\text{g}\dfrac{d\text{f}}{dx}+\text{f}\dfrac{d\text{g}}{dx}$
3) Quotient rule:
$\dfrac{d}{dx}\left( \dfrac{\text{f}}{\text{g}} \right)=\dfrac{\text{g}\dfrac{d\text{f}}{dx}-\text{f}\dfrac{d\text{g}}{dx}}{{{\text{g}}^{\text{2}}}}$