Question

lim Cos2x-1
x--->0 ------------
Cosx-1 ​

Answer 1

$\large\underline{\sf{Solution-}}$

$\sf \:\: \: \: \displaystyle\sf \:\lim_{x\to 0} \: \dfrac{cos2x - 1}{cosx - 1}$

$= \: \: \: \displaystyle\sf \:\lim_{x\to 0}\dfrac{1 - cos2x}{1 - cosx}$

$\sf \: = \: \: \: \dfrac{1 - cos0}{1 - cos0}$

$\sf \: = \: \: \: \dfrac{1 - 1}{1 - 1}$

$\sf \: = \: \: \: \dfrac{0}{0}$

which is indeterminant form.

Hence,

To solve this limit, we break down the formula,

$\boxed{ \sf\ \: 1 - cosx = 2 {sin}^{2}\dfrac{x}{2}} \: and \: \boxed{ \sf \: 1 - cos2x = 2 {sin}^{2} x}$

So,

$\: \: \: \displaystyle\sf \:\lim_{x\to 0}\dfrac{1 - cos2x}{1 - cosx}$

$= \: \: \: \displaystyle\sf \:\lim_{x\to 0}\dfrac{ \cancel2 \: {sin}^{2}x }{ \cancel2 \: {sin}^{2}\dfrac{x}{2} }$

$= \: \: \: \displaystyle\sf \:\lim_{x\to 0}\dfrac{sinx \: sinx}{sin\dfrac{x}{2} \: sin\dfrac{x}{2}}$

$= \: \: \: \displaystyle\sf \:\lim_{x\to 0}\dfrac{\bigg(2 \: \cancel{sin\dfrac{x}{2}} \: cos\dfrac{x}{2} \bigg) \bigg(2 \: \cancel{sin\dfrac{x}{2}} \: cos\dfrac{x}{2} \bigg) }{ \cancel{sin\dfrac{x}{2}} \: \: \cancel{sin\dfrac{x}{2}}}$

$= \: \: \: 4 \times \displaystyle\sf \:\lim_{x\to 0}cos\dfrac{x}{2} \: \times \: \: \: \: \displaystyle\sf \:\lim_{x\to 0}cos\dfrac{x}{2}$

$\sf \: = \: \: \: 4 \times 1 \times 1$

$\sf \: = \: \: \: 4$

Additional Information :-

$\: \: \: \displaystyle\sf \:\lim_{x\to 0}\dfrac{sinx}{x} = 1$

$\: \: \: \displaystyle\sf \:\lim_{x\to 0}\dfrac{tanx}{x} = 1$

$\: \: \: \displaystyle\sf \:\lim_{x\to 0}\dfrac{ {e}^{x} - 1}{x} = 1$

$\: \: \: \displaystyle\sf \:\lim_{x\to 0}\dfrac{ log(1 + x) }{x} = 1$

$\: \: \: \displaystyle\sf \:\lim_{x\to 0}\dfrac{ {a}^{x} - 1}{x} = {a}^{x} log(a)$