Question

Solve $ {\lim _{x \Rightarrow 0}}\dfrac{{\sqrt {2 + x} - \sqrt 2 }}{x} $

Answer 1

Hint: For finding the value of the given expression, we can substitute the given value of limit. But if we get an indeterminate form after the substitution, we can then use L'hospital's rule. In this, we take the derivatives individually of the term in numerator and denominator and then substitute the value of limit.
L'hospital's rule states that:
${\text{If }}{\lim _{x \Rightarrow 0}}\dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{0}{0}{\text{ or }}\dfrac{\infty }{\infty } \\
  {\text{Then, }}{\lim _{x \Rightarrow 0}}\dfrac{{f\left( x \right)}}{{g\left( x \right)}} = {\lim _{x \Rightarrow 0}}\dfrac{{f'\left( x \right)}}{{g'\left( x \right)}} \\
  $
To calculate the derivative remember:
 $ \dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} $

Complete step-by-step answer:
The given mathematical expression is $ {\lim _{x \Rightarrow 0}}\dfrac{{\sqrt {2 + x} - \sqrt 2 }}{x} $ .
The limit of the variable x is approaching 0. Its value can be determined by substituting this value of limit in the expression.
Substituting x = 0 in the given mathematical expression:
 $
  \dfrac{{\sqrt {2 + 0} - \sqrt 2 }}{0} \Rightarrow \dfrac{{\sqrt 2 - \sqrt 2 }}{0} \\
   \Rightarrow \dfrac{0}{0} \;
  $
As after substituting the limit, no determinate value is obtained. This form is called as indeterminate form $ \dfrac{0}{0} $
To evaluate the limits in such cases, L'hospital's rule is used. It states that:
 $
{\text{If }}{\lim _{x \Rightarrow 0}}\dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{0}{0}{\text{ or }}\dfrac{\infty }{\infty } \\
  {\text{Then, }}{\lim _{x \Rightarrow 0}}\dfrac{{f\left( x \right)}}{{g\left( x \right)}} = {\lim _{x \Rightarrow 0}}\dfrac{{f'\left( x \right)}}{{g'\left( x \right)}} \;
  $
For the given expression,
 $
  f(x) = \sqrt {2 + x} - \sqrt 2 \\
  g(x) = x \;
  $
Their derivatives are given as:
\[
  f'(x) = \dfrac{d}{{dx}}\left( {\sqrt {2 + x} - \sqrt 2 } \right) \\
   \Rightarrow \dfrac{d}{{dx}}\left[ {{{\left( {2 + x} \right)}^{\dfrac{1}{2}}} - 0} \right] \\
   \Rightarrow f'(x) = \dfrac{1}{{2\sqrt {2 + x} }} \\
  g'(x) = \dfrac{d}{{dx}}\left( x \right) \\
   \Rightarrow g'(x) = 1 \;
 \]
And the value we obtained was:
 $ {\lim _{x \Rightarrow 0}}\dfrac{{\sqrt {2 + x} - \sqrt 2 }}{x} = \dfrac{0}{0} $
Applying L hospital’s rule and substituting the values, we get:
 $
  {\lim _{x \Rightarrow 0}}\dfrac{{f\left( x \right)}}{{g\left( x \right)}} = {\lim _{x \Rightarrow 0}}\dfrac{{f'\left( x \right)}}{{g'\left( x \right)}} \\
   \Rightarrow {\lim _{x \Rightarrow 0}}\dfrac{{f'\left( x \right)}}{{g'\left( x \right)}} = {\lim _{x \Rightarrow 0}}\dfrac{{\dfrac{1}{{2\sqrt {2 + x} }}}}{1} \;
  $
Now substituting x = 0, we get:
 $
   \Rightarrow {\lim _{x \Rightarrow 0}}\dfrac{{f'\left( x \right)}}{{g'\left( x \right)}} = \dfrac{{\dfrac{1}{{2\sqrt {2 + 0} }}}}{1} \\
   \Rightarrow {\lim _{x \Rightarrow 0}}\dfrac{{f'\left( x \right)}}{{g'\left( x \right)}} = \dfrac{1}{{2\sqrt 2 }} \;
  $
And
 $
  {\lim _{x \Rightarrow 0}}\dfrac{{f\left( x \right)}}{{g\left( x \right)}} = {\lim _{x \Rightarrow 0}}\dfrac{{f'\left( x \right)}}{{g'\left( x \right)}} \\
   \Rightarrow {\lim _{x \Rightarrow 0}}\dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{1}{{2\sqrt 2 }} \\
  $
Therefore, the value the given expression $ {\lim _{x \Rightarrow 0}}\dfrac{{\sqrt {2 + x} - \sqrt 2 }}{x} $ is $ \dfrac{1}{{2\sqrt 2 }} $
So, the correct answer is “$ \dfrac{1}{{2\sqrt 2 }} $”.

Note: We can obtain indeterminate form sometimes even after applying L'hospital's rule, so it can be applied more than once until we get out of indeterminate form and obtain a determinate value. When the limits are applied to a function and it takes indeterminate form, then we can apply L hospital’s rule which is applicable to all kinds of indeterminate forms. Some common indeterminate forms are:
\[\dfrac{0}{0}{\text{, }}\dfrac{\infty }{\infty },{1^0}{\text{, }}{0^0}\]