Question

Evaluate limit x tends to 1 (e^x - 1)÷root over 1 - cosx

Answer 1

Please see the attachment

Answer 2

The correct question is

Evaluate limit x tends to 0 for f(x) = $\frac{e^{x}-1 }{\sqrt{1-cosx} }$  

The solution of  

$\lim_{x \to 0\1} \frac{e^{x}-1 }{\sqrt{1-cosx} } = \sqrt{2}$  

• We have,

      $\lim_{x \to 0\1} \frac{e^{x}-1 }{\sqrt{1-cosx} }$

• Now, by using double angle formula for cosx that is

         $cos2x = 1 - 2sin^{2}x\\ 1 - cos2x= 2sin^{2}x$

      we get,

             $=\lim_{x \to 0\1} \frac{e^{x}-1 }{\sqrt{2sin^{2}\frac{x}{2} } }$

             $=\lim_{x \to 0\1} \frac{e^{x}-1 }{\sqrt{2} sin\frac{x}{2} }$

• dividing both numerator and denominator by 'x' , we get

       $= \lim_{x \to 0\1} \frac{\frac{e^{x}-1 }{x} }\sqrt{2} {\frac{sin\frac{x}{2} }{\frac{x}{2}2 } }$

       $= \frac{2}{\sqrt{2} } \frac{ \lim_{x \to 0\1} \frac{e^{x}-1 }{x} }{ \lim_{x \to 0\1} \frac{sin\frac{x}{2} }{\frac{x}{2} } }$

• Now, by using standard sine limit and exponential limits that is

     $\lim_{x \to 0\1} \frac{sinx}{x} = 1 and \lim_{x \to 0\1} \frac{e^{x} -1}{x} = 1$ , we get

     $\frac{2}{\sqrt{2} } \frac{ \lim_{x \to 0\1} \frac{e^{x}-1 }{x} }{ \lim_{x \to 0\1} \frac{sin\frac{x}{2} }{\frac{x}{2} } } = \frac{2}{\sqrt{2} } (\frac{1}{1} )$

Therefore,

$\lim_{x \to 0\1} \frac{e^{x}-1 }{\sqrt{1-cosx} } = \sqrt{2}$