Question

Evaluate the following

$lim \: x \to \: 0 \: \frac{cos3x - cosx}{cos7x - cos5x}$
​

Answer 1

Step-by-step explanation:

We have,

$\displaystyle \tt{ \lim_{x \rarr0 } \: \dfrac{ cos(3x) - cos(x) }{ cos(7x) - cos(5x) } }$

$\displaystyle \tt{ = \lim_{x \rarr0 } \: \dfrac{ - 2 \: sin \left( \dfrac{3x + x}{2} \right) sin \left( \dfrac{3x - x}{2} \right) }{ - 2 \: sin \left( \dfrac{7x +5x}{2} \right) sin \left( \dfrac{7x -5 x}{2} \right) } }$

$\displaystyle \tt{ = \lim_{x \rarr0 } \: \dfrac{ sin \left( \dfrac{4x}{2} \right) sin \left( \dfrac{2x }{2} \right) }{ sin \left( \dfrac{12x}{2} \right) sin \left( \dfrac{2x}{2} \right) } }$

$\displaystyle \tt{ = \lim_{x \rarr0 } \: \dfrac{ sin \left(2x \right) sin \left(x\right) }{ sin \left( 6x \right) sin \left(x \right) } }$

$\displaystyle \tt{ = \lim_{x \rarr0 } \: \dfrac{ sin \left(2x \right) }{ sin \left( 6x \right) } }$

$\displaystyle \tt{ = \lim_{x \rarr0 } \: \dfrac{ sin \left(2x \right) }{x} \cdot\dfrac{x }{ sin \left( 6x \right) } }$

$\displaystyle \tt{ = \lim_{x \rarr0 } \: \dfrac{ 2 \cdot sin \left(2x \right) }{2x} \cdot\dfrac{6x }{6 \cdot sin \left( 6x \right) } }$

$\displaystyle \tt{ = \dfrac{2}{6} \: \lim_{x \rarr0 } \: \dfrac{sin \left(2x \right) }{2x} \cdot \lim _{x \to0}\dfrac{6x }{ sin \left( 6x \right) } }$

$\displaystyle \tt{ = \dfrac{2}{6} \cdot1 \cdot1 }$

$\displaystyle \tt{ = \dfrac{1}{3} }$

Answer 2

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{cos3x - cosx}{cos7x - cos5x}$

If we substitute directly x = 0, we get

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

$\rm \:  =  \: \dfrac{1 - 1}{1 - 1}$

$\rm \:  =  \: \dfrac{0}{0}$

which is indeterminant form.

So, Consider again

$\rm :\longmapsto\:\displaystyle\lim_{x \to 0}\rm \frac{cos3x - cosx}{cos7x - cos5x}$

We know, By L - Hospital Rule we have

$\rm \:  =  \: \:\displaystyle\lim_{x \to 0}\rm \frac{\dfrac{d}{dx}cos3x - \dfrac{d}{dx}cosx}{\dfrac{d}{dx}cos7x -\dfrac{d}{dx} cos5x}$

We know,

$\rm :\longmapsto\:\boxed{\tt{ \dfrac{d}{dx}cosx \: = \: - \: sinx}}$

So, using this, we get

$\rm \:  =  \: \displaystyle\lim_{x \to 0}\rm \frac{ - sin3x \times 3 + sinx}{ - sin7x \times 7 + sin5x \times 5}$

$\rm \:  =  \: \displaystyle\lim_{x \to 0}\rm \frac{ - 3sin3x + sinx}{ - 7sin7x + 5sin5x }$

If we substitute x = 0, we again get indeterminant form.

So, again using L - Hospital Rule, we get

$\rm \:  =  \: \displaystyle\lim_{x \to 0}\rm \frac{ - 3\dfrac{d}{dx}sin3x + \dfrac{d}{dx}sinx}{ - 7\dfrac{d}{dx}sin7x + 5\dfrac{d}{dx}sin5x }$

$\rm \:  =  \: \displaystyle\lim_{x \to 0}\rm \frac{ - 9cos3x + cosx}{ - 49cos7x + 25cos5x}$

$\rm \:  =  \: \dfrac{ - 9cos0 + cos0}{ - 49cos0 + 25cos0}$

$\rm \:  =  \: \dfrac{ - 9 + 1}{ - 49 + 25}$

$\rm \:  =  \: \dfrac{ - 8}{ - 24}$

$\rm \:  =  \: \dfrac{1}{3}$

Hence,

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{cos3x - cosx}{cos7x - cos5x} = \frac{1}{3}}}}$

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More to Know

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$