Question

Without using L Hospital Rule evaluate

$\lim \: x \to \: \frac{\pi}{3} \: \: \: \: \frac{2 \: sin(x - \frac{\pi}{3}) }{1 - 2 \: cosx}$
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Answer 1

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

Given expression is

$\rm \: \displaystyle\lim_{x \to \dfrac{\pi}{3} }\rm \: \frac{2sin\bigg(x - \dfrac{\pi}{3} \bigg) }{1 - 2cosx}$

If we substitute directly the value of x, we get

$\rm \: =  \: \dfrac{2sin0}{1 - 2cos\dfrac{\pi}{3} }$

$\rm \: =  \: \dfrac{0}{1 - 2 \times \dfrac{1}{2} }$

$\rm \: =  \: \dfrac{0}{1 - 1}$

$\rm \: =  \: \dfrac{0}{0}$

which is indeterminant form.

So, Consider again

$\rm \: \displaystyle\lim_{x \to \dfrac{\pi}{3} }\rm \: \frac{2sin\bigg(x - \dfrac{\pi}{3} \bigg) }{1 - 2cosx}$

To evaluate this limit, we use method of Substitution.

So, Substitute

$\rm \: x = \dfrac{\pi}{3} + h, \: \: as \: x \: \to \: \dfrac{\pi}{3} , \: \: so \: h \: \to \: 0$

So, on substituting these values, we get

$\rm \:=  \: 2 \displaystyle\lim_{h \to 0 }\rm \: \frac{sin\bigg(\dfrac{\pi}{3} + h - \dfrac{\pi}{3} \bigg) }{1 - 2cos\bigg(\dfrac{\pi}{3} + h\bigg) }$

$\rm \:=  \: 2\displaystyle\lim_{h \to 0 }\rm \: \frac{sin\bigg(h\bigg) }{1 - 2\bigg(cos\dfrac{\pi}{3}cosh - sin\dfrac{\pi}{3} sinh\bigg) }$

$\rm \:=  \: 2 \displaystyle\lim_{h \to 0 }\rm \: \frac{sin\bigg(h\bigg) }{1 - 2\bigg( \dfrac{1}{2} cosh - \dfrac{ \sqrt{3} }{2} sinh\bigg) }$

$\rm \:=  \:2 \displaystyle\lim_{h \to 0 }\rm \: \frac{sinh }{1 - cosh + \sqrt{3} sinh}$

$\rm \:=  \: 2\displaystyle\lim_{h \to 0 }\rm \: \frac{2sin \dfrac{h}{2}cos\dfrac{h}{2} }{ {2sin}^{2}\dfrac{h}{2} + 2\sqrt{3}sin\dfrac{h}{2}cos\dfrac{h}{2}}$

$\rm \:=  \: 2\displaystyle\lim_{h \to 0 }\rm \: \frac{2sin \dfrac{h}{2}cos\dfrac{h}{2} }{ {2sin}\dfrac{h}{2}\bigg(sin\dfrac{h}{2} + \sqrt{3} \: cos\dfrac{h}{2}\bigg)}$

$\rm \:=  \: 2 \displaystyle\lim_{h \to 0 }\rm \: \frac{cos\dfrac{h}{2} }{ sin\dfrac{h}{2} + \sqrt{3} \: cos\dfrac{h}{2}}$

$\rm \: =  \: \dfrac{2}{0 + \sqrt{3} }$

$\rm \: =  \: \dfrac{2}{ \sqrt{3} }$

Hence,

$\rm\implies \:\boxed{\sf{  \: \: \rm \: \displaystyle\lim_{x \to \dfrac{\pi}{3} }\rm \: \frac{sin\bigg(x - \dfrac{\pi}{3} \bigg) }{1 - 2cosx} = \frac{2}{ \sqrt{3} } \: }}$

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Formulae Used :-

$\boxed{\rm{  \:cos(x + y) = cosxcosy - sinxsiny \: }}$

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

$\boxed{\rm{  \:sin2x = 2 \: sinx \: cosx \: }}$

$\boxed{\sf{  \:cos\dfrac{\pi}{3} = \frac{1}{2} \: }}$

$\boxed{\sf{  \:sin\dfrac{\pi}{3} = \frac{ \sqrt{3} }{2} \: }}$

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Additional Information :-

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

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

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

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

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