Question

\(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sqrt {\left( {\frac{1}{2}\left( {1 - {\rm{cos}}2{\rm{x}}} \right)} \right)} }}{{\rm{x}}}\) is equal to
1. 1
2. -1
3. 0
4. 2
5. none of these

Answer 1

Correct Answer - Option 5 : none of these

Concept:

• Limit exists if and only if LHL = RHL

i.e.,

• (√x²) = |x|
• \({\rm{f}}\left( {\rm{x}} \right) = \left| {\rm{x}} \right| = \left\{ {\begin{array}{*{20}{c}} { - x,\;x < 0}\\ {x,\;x \ge 0} \end{array}} \right.\)

• cos2x = 1 – 2sin²x

 

Calculation:

To find: \(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sqrt {\left( {\frac{1}{2}\left( {1 - {\rm{cos}}2{\rm{x}}} \right)} \right)} }}{{\rm{x}}} = ?\)

\(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sqrt {\left( {\frac{1}{2}\left( {1 - {\rm{cos}}2{\rm{x}}} \right)} \right)} }}{{\rm{x}}} = \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sqrt {\left( {\frac{1}{2}(2{{\sin }^2}{\rm{x}})} \right)} }}{{\rm{x}}}\)

\(= \mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{\sqrt {\left( {{{\sin }^2}{\rm{x}}} \right)} }}{{\rm{x}}}\)

\(= \mathop {{\rm{lim}}}\limits_{{\rm{x}} \to 0} \frac{{\left| {{\rm{sinx}}} \right|}}{{\rm{x}}}\)                         (∵ (√x²) = |x|)

Now, LHL = \(\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {0^ - }} \frac{{\left| {{\rm{sinx}}} \right|}}{{\rm{x}}} = {\rm{\;}}\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {0^ - }} \frac{{ - {\rm{sinx}}}}{{\rm{x}}} = - 1\)

RHL = \(\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {0^ + }} \frac{{\left| {{\rm{sinx}}} \right|}}{{\rm{x}}} = {\rm{\;}}\mathop {{\rm{lim}}}\limits_{{\rm{x}} \to {0^ - }} \frac{{{\rm{sinx}}}}{{\rm{x}}} = 1\)

Here LHL ≠ RHL, so limit doesn’t exist.

Hence, option (5) is correct.