Question

The value of \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 - \cos x} }}{x}\] is equal to
A.\[ - \dfrac{1}{{\sqrt 2 }}\]
B.\[\dfrac{1}{{\sqrt 2 }}\]
C.\[0\]
D.Does not exist

Answer 1

Hint: Here in this question, we have to determine the given limit of a function. To find this first we have to write the given function using a trigonometric double or half angle formula \[\cos 2x = 1 - 2{\sin ^2}x\] then limit of a function \[f\] Which is satisfies the condition left hand limit is equal to right hand limit (i.e., \[LHL = RHL\]) by applying a limit in to the function using the properties of limits, otherwise limits doesn’t exist

Complete step by step solution:
The limit of a function exists if and only if the left-hand limit is equal to the right-hand limit.
A left-hand limit means the limit of a function as it approaches from the left-hand side.
\[\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = {l_1}\]
A right-hand limit means the limit of a function as it approaches from the right-hand side.
\[\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = {l_2}\]
Consider the given limit function,
\[ \Rightarrow \,\,\,\mathop {\lim }\limits_{x \to 0} \dfrac{{\sqrt {1 - \cos x} }}{x}\]------(1)
By using a double of half angle formula: \[\cos 2x = 1 - 2{\sin ^2}x \Rightarrow 1 - \cos x = 2{\sin ^2}\dfrac{x}{2}\], then
As we know, \[\sqrt {1 - \cos x} = \left\{ {\begin{array}{*{20}{c}}
  { - \sqrt 2 \sin \dfrac{x}{2},\,\,\,\,\,\,\,x < 0} \\
  {\,\,\sqrt 2 \sin \dfrac{x}{2},\,\,\,\,\,\,\,x \geqslant 0}
\end{array}} \right.\]
Now, find the left-hand limit to the function (1)
\[ \Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\sqrt {1 - \cos x} }}{x}\]
\[ \Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} - \dfrac{{\sqrt 2 \sin \dfrac{x}{2}}}{x}\]
By applying a properties of limit function, we have
\[ \Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - \sqrt 2 \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\sin \dfrac{x}{2}}}{x}\]
Multiply and divide by \[\dfrac{1}{2}\], then
\[ \Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - \sqrt 2 \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\sin \dfrac{x}{2}}}{x} \times \dfrac{{\left( {\dfrac{1}{2}} \right)}}{{\left( {\dfrac{1}{2}} \right)}}\]
\[ \Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - \sqrt 2 \mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\left( {\dfrac{1}{2}} \right)\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}\]
Again, by the properties of limit, we have
\[ \Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - \sqrt 2 \left( {\dfrac{1}{2}} \right)\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}\]
As we know the standard limit \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1\], then
\[ \Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - \dfrac{{\sqrt 2 }}{2}\]
\[ \Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = - \dfrac{1}{{\sqrt 2 }}\]---------(2)
Now, find the right-hand limit to the function (1)
\[ \Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sqrt {1 - \cos x} }}{x}\]
\[ \Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sqrt 2 \sin \dfrac{x}{2}}}{x}\]
By applying a properties of limit function, we have
\[ \Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \sqrt 2 \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sin \dfrac{x}{2}}}{x}\]
Multiply and divide by \[\dfrac{1}{2}\], then
\[ \Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = - \sqrt 2 \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sin \dfrac{x}{2}}}{x} \times \dfrac{{\left( {\dfrac{1}{2}} \right)}}{{\left( {\dfrac{1}{2}} \right)}}\]
\[ \Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = - \sqrt 2 \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\left( {\dfrac{1}{2}} \right)\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}\]
Again, by the properties of limit, we have
\[ \Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = - \sqrt 2 \left( {\dfrac{1}{2}} \right)\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sin \dfrac{x}{2}}}{{\dfrac{x}{2}}}\]
As we know the standard limit \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1\], then
\[ \Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = - \dfrac{{\sqrt 2 }}{2}\]
\[ \Rightarrow \,\,\,\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \dfrac{1}{{\sqrt 2 }}\]---------(3)
Since, by (2) and (3)
\[\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{{\sqrt {1 - \cos x} }}{x} \ne \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\sqrt {1 - \cos x} }}{x}\]
\[LHL \ne RHL\]
Hence, limit does not exist
Therefore, option (D) is correct
So, the correct answer is “Option D”.

Note: Remember, the limit of any function exists between any two consecutive integers. And the product and quotient properties of limits are defined as:
The function \[f\left( x \right)\] and \[g\left( x \right)\] is are non-zero finite values, given that
\[\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) \cdot g\left( x \right)} \right) = \mathop {\lim }\limits_{x \to a} f\left( x \right) \cdot \mathop {\lim }\limits_{x \to a} g\left( x \right)\]
\[\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \dfrac{{\mathop {\lim }\limits_{x \to a} f\left( x \right)}}{{\mathop {\lim }\limits_{x \to a} g\left( x \right)}}\] and
Also \[\mathop {\lim }\limits_{x \to a} k\,f\left( a \right) = k\mathop {\lim }\limits_{x \to a} f\left( a \right)\].