Question

lim x tends to π/6[2-√3cos x- sin x/(6x-π)²]​

Answer 1

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \bf \: sin(x + y) = sinxcosy + cosxsiny}$

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

$\boxed{ \bf \:1 - cosx = 2 {sin}^{2} \dfrac{x}{2} }$

$\boxed{ \bf \: \tt \:\lim_{h\to0}\dfrac{sinh}{h} = 1}$

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

$\tt \:\lim_{x\to \frac{\pi}{6} }\dfrac{2 - \sqrt{3}cosx - sinx}{ {(6x - \pi \:)}^{2} }$

On directly Substitute the value of x, we get indeterminant form.

So, to solve such questions, we prefer Substitution method.

So,

$\tt \: Put \: x = \dfrac{\pi \:}{6} + h \: \: \: As \: x \to \: \dfrac{\pi}{6} \: so \: h \to \: 0$

$\tt = \:\lim_{h\to0}\dfrac{2 - \sqrt{3}cos\bigg(\dfrac{\pi \:}{6} + h\bigg) - sin\bigg( \dfrac{\pi \:}{6} + h \bigg) }{ {\bigg(6(\dfrac{\pi \:}{6} + h) - \pi \: \bigg) }^{2} }$

$\tt = \:\lim_{h\to0}\dfrac{2 - \sqrt{3}cos\bigg(\dfrac{\pi \:}{6} + h\bigg) - sin\bigg( \dfrac{\pi \:}{6} + h \bigg) }{ {\bigg( \pi \:+6 h - \pi \: \bigg) }^{2} }$

$\tt \: = \lim_{h\to0}\dfrac{2 - \sqrt{3}cos\bigg(\dfrac{\pi \:}{6} + h\bigg) - sin\bigg( \dfrac{\pi \:}{6} + h \bigg) }{ {\bigg( 6 h \bigg) }^{2} }$

$\tt = \:\lim_{h\to0}\dfrac{2 - \sqrt{3}cos\bigg(\dfrac{\pi \:}{6} + h\bigg) - sin\bigg( \dfrac{\pi \:}{6} + h \bigg) }{ {36 {h}^{2} }} - - (1)$

Consider,

$\tt \: \sqrt{3} \: cos\bigg(\dfrac{\pi \:}{6} + h \bigg)$

$= \tt \: \sqrt{3} \: (cos\dfrac{\pi \:}{6} cosh - sin\dfrac{\pi \:}{6} sinh)$

$= \tt \: \sqrt{3} \: (\dfrac{ \sqrt{3} }{2}cosh - \dfrac{1}{2} sinh)$

$= \tt \: \dfrac{3}{2}cosh - \dfrac{ \sqrt{3} }{2} sinh$

Again,

Consider,

$\tt \: sin\bigg(\dfrac{\pi \:}{6} + h \bigg)$

$= \tt \: sin\dfrac{\pi \:}{6} cosh + cos\dfrac{\pi \:}{6} sinh$

$= \tt \: \dfrac{1}{2} cosh + \dfrac{ \sqrt{3} }{2} sinh$

On substituting these values in equation (1), we get

$\tt \:\lim_{h\to0}\dfrac{2 - \dfrac{ 3}{2} cosh + \cancel{\dfrac{ \sqrt{3} }{2}sinh} - \dfrac{1}{2} cosh -\cancel {\dfrac{ \sqrt{3} }{2}sinh}}{36 {h}^{2} }$

$\tt \:\lim_{h\to0}\dfrac{2 - 2cosh}{36 {h}^{2} }$

$\tt \:\lim_{h\to0}\dfrac{1 - cosh}{18 {h}^{2} }$

$\tt \:\lim_{h\to0}\dfrac{2 {sin}^{2} \dfrac{h}{2} }{18 {h}^{2} }$

$\tt \:\dfrac{1}{9} \lim_{h\to0}\dfrac{sin\dfrac{h}{2} \times sin \dfrac{h}{2} }{ {h}^{2} }$

$\tt \: =\dfrac{1}{9} \: \lim_{h\to0} \dfrac{sin\dfrac{h}{2} }{\dfrac{h}{2} \times 2 } \times \dfrac{sin\dfrac{h}{2} }{\dfrac{h}{2} \times 2 }$

$\tt \: = \: \dfrac{1}{9 \times 2 \times 2}$

$\tt \: = \dfrac{1}{36}$

Additional Information :-

$\boxed{ \bf \: \tt \:\lim_{h\to0}\dfrac{tanh}{h} = 1}$

$\boxed{ \bf \: \tt \:\lim_{h\to0}\dfrac{ {e}^{h} - 1}{h} = 1}$

$\boxed{ \bf \: \tt \:\lim_{h\to0}\dfrac{ {a}^{h} - 1}{h} = log(a) }$

$\boxed{ \bf \: \tt \:\lim_{h\to0}\dfrac{ log(1 + h) }{h} = 1}$

$\boxed{ \bf \: \tt \:\lim_{x\to \: a}\dfrac{ {x}^{n} - {a}^{n} }{x - a} = {na}^{n - 1} }$