Question

Evaluate the following

$lim \: x \to \: 0 \: \frac{cosx - cos2x}{1 - cosx}$
​

Answer 1

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{cosx - cos2x}{1 - cosx}$

If we substitute directly x = 0, we get

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

$\rm \:  =  \: \dfrac{1 - 1}{1 - 1}$

$\rm \:  =  \: \dfrac{0}{0}$

which is indeterminant form.

So, Consider

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{cosx - cos2x}{1 - cosx}$

We know,

$\boxed{\tt{ cosx - cosy = 2sin\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{y - x}{2} \bigg]}}$

And

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

So, using these Identities, we get

$\rm \:  =  \: \displaystyle\lim_{x \to 0}\rm \frac{2sin\bigg[\dfrac{x + 2x}{2} \bigg]sin\bigg[\dfrac{2x - x}{2} \bigg]}{ {2sin}^{2} \dfrac{x}{2} }$

$\rm \:  =  \: \displaystyle\lim_{x \to 0}\rm \frac{sin\bigg[\dfrac{3x}{2} \bigg]sin\bigg[\dfrac{x}{2} \bigg]}{ {sin}^{2} \dfrac{x}{2} }$

$\rm \:  =  \: \displaystyle\lim_{x \to 0}\rm \frac{sin\bigg[\dfrac{3x}{2} \bigg]}{sin \dfrac{x}{2} }$

can be rewritten as

$\rm \:  =  \: \displaystyle\lim_{x \to 0}\rm \dfrac{sin\bigg[\dfrac{3x}{2} \bigg]}{\dfrac{3x}{2} } \times \dfrac{3x}{2} \times \dfrac{\dfrac{x}{2}}{sin\dfrac{x}{2} } \times \dfrac{2}{x}$

We know,

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \: \frac{sinx}{x} = 1 \: }}$

So, using this identity, we get

$\rm \:  =  \: 1 \times 3 \times 1$

$\rm \:  =  \: 3$

Hence,

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{cosx - cos2x}{1 - cosx} = 3}}}$

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More to Know

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \: \frac{sinx}{x} = 1 \: }}$

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \: \frac{tanx}{x} = 1 \: }}$

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \: \frac{log(1 + x)}{x} = 1 \: }}$

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \: \frac{ {e}^{x} - 1}{x} = 1 \: }}$

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \: \frac{ {a}^{x} - 1}{x} = loga \: }}$

Answer 2

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{cosx - cos2x}{1 - cosx}$

If we substitute directly x = 0, we get

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

$\rm \:  =  \: \dfrac{1 - 1}{1 - 1}$

$\rm \:  =  \: \dfrac{0}{0}$

which is indeterminant form.

So, Consider

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{cosx - cos2x}{1 - cosx}$

We know,

$\boxed{\tt{ cosx - cosy = 2sin\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{y - x}{2} \bigg]}}$

And

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

So, using these Identities, we get

$\rm \:  =  \: \displaystyle\lim_{x \to 0}\rm \frac{2sin\bigg[\dfrac{x + 2x}{2} \bigg]sin\bigg[\dfrac{2x - x}{2} \bigg]}{ {2sin}^{2} \dfrac{x}{2} }$

$\rm \:  =  \: \displaystyle\lim_{x \to 0}\rm \frac{sin\bigg[\dfrac{3x}{2} \bigg]sin\bigg[\dfrac{x}{2} \bigg]}{ {sin}^{2} \dfrac{x}{2} }$

$\rm \:  =  \: \displaystyle\lim_{x \to 0}\rm \frac{sin\bigg[\dfrac{3x}{2} \bigg]}{sin \dfrac{x}{2} }$

can be rewritten as

$\rm \:  =  \: \displaystyle\lim_{x \to 0}\rm \dfrac{sin\bigg[\dfrac{3x}{2} \bigg]}{\dfrac{3x}{2} } \times \dfrac{3x}{2} \times \dfrac{\dfrac{x}{2}}{sin\dfrac{x}{2} } \times \dfrac{2}{x}$

We know,

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \: \frac{sinx}{x} = 1 \: }}$

So, using this identity, we get

$\rm \:  =  \: 1 \times 3 \times 1$

$\rm \:  =  \: 3$

Hence,

$\green{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{cosx - cos2x}{1 - cosx} = 3}}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

More to Know

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \: \frac{sinx}{x} = 1 \: }}$

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \: \frac{tanx}{x} = 1 \: }}$

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \: \frac{log(1 + x)}{x} = 1 \: }}$

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \: \frac{ {e}^{x} - 1}{x} = 1 \: }}$

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \: \frac{ {a}^{x} - 1}{x} = loga \: }}$