Question

lim
3
2-13 cos x-sin x
(6x-7)
2
π
X->
6
plz ans it correct step by step!​

Answer 1

Basic Identities Used :-

$\begin{gathered}(1)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{sin \: x}{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(2)\:{\underline{\boxed{\bf{\blue{{\tt \: sin(x + y) = sinx \: cosy + siny \: cosx}}}}}} \\ \end{gathered}$

$\begin{gathered}(3)\:{\underline{\boxed{\bf{\blue{{\tt \: cos(x + y) = cosx \: cosy - siny \: sinx}}}}}} \\ \end{gathered}$

$\begin{gathered}(4)\:{\underline{\boxed{\bf{\blue{{\tt \: sin\dfrac{\pi}{6} = \dfrac{1}{2} }}}}}} \\ \end{gathered}$

$\begin{gathered}(5)\:{\underline{\boxed{\bf{\blue{{\tt \: cos\dfrac{\pi}{6} = \dfrac{ \sqrt{3} }{2} }}}}}} \\ \end{gathered}$

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

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

If we put directly the value of x, we have

$\rm \: = \: \dfrac{2 - \sqrt{3}cos \frac{\pi}{6} - sin \frac{\pi}{6} }{ {(6 \times \frac{\pi}{6} - \pi)}^{2} }$

$\rm \: = \: \dfrac{2 - \sqrt{3} \times \dfrac{ \sqrt{3} }{2} - \dfrac{1}{2} }{(\pi - \pi)^{2} }$

$\rm \: = \: \dfrac{0 }{(0)^{2} }$

$\rm \: = \: \dfrac{0 }{(0)}$

which is indeterminant form.

So,

We solve the limit by substitution method

Consider,

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

$\red{\bf :\longmapsto\:Put \: x = \dfrac{\pi}{6} + h}$

$\red{\bf :\longmapsto\:As \: x \to \dfrac{\pi}{6} \: \: \: so \: \: \: h \to \: 0}$

Given limit can be rewritten as

$= \: \displaystyle\bf \:\lim_{h\to 0} \: \dfrac{2 - \sqrt{3}cos\bigg(\dfrac{\pi}{6} + h\bigg) - sin\bigg(\dfrac{\pi}{6} + h\bigg) }{ { \bigg(6\bigg(\dfrac{\pi}{6} + h\bigg) - \pi \bigg)}^{2} }$

$\displaystyle\sf = \:\lim_{h\to 0} \: \dfrac{2 - \sqrt{3}\bigg(cos\frac{\pi}{6}cosh - sin \frac{\pi}{6}sinh\bigg) - \bigg(sin\frac{\pi}{6}cosh + sinhcos \frac{\pi}{6} \bigg) }{ { \bigg(6h\bigg)}^{2} }$

$\displaystyle\sf = \:\lim_{h\to 0} \: \dfrac{2 - \sqrt{3}\bigg(\frac{ \sqrt{3} }{2}cosh -\frac{1}{2}sinh\bigg) - \bigg(\frac{1}{2}cosh + sinh\frac{ \sqrt{3} }{2} \bigg) }{ { \bigg(6h\bigg)}^{2}}$

$\displaystyle\sf = \:\lim_{h\to 0} \: \dfrac{2 - \bigg(\frac{3}{2}cosh -\frac{ \sqrt{3} }{2}sinh\bigg) - \bigg(\frac{1}{2}cosh + sinh\frac{ \sqrt{3} }{2} \bigg) }{ {36h}^{2}}$

$\displaystyle\sf = \:\lim_{h\to 0} \: \dfrac{2 - \frac{3}{2}cosh + \cancel{ \frac{ \sqrt{3} }{2}sinh} -\frac{1}{2}cosh - \cancel{sinh\frac{ \sqrt{3} }{2}}}{ {36h}^{2}}$

$\displaystyle\sf = \:\lim_{h\to 0} \: \dfrac{2 - 2cosh}{ {36h}^{2}}$

$\displaystyle\sf = \:\lim_{h\to 0} \: \dfrac{1 - cosh}{ {18h}^{2}}$

$\displaystyle\sf = \:\lim_{h\to 0} \: \dfrac{2 {sin}^{2}\dfrac{h}{2} }{ {18h}^{2}}$

$\displaystyle\sf = \:\lim_{h\to 0} \: \dfrac{ {sin}\dfrac{h}{2} \times sin\dfrac{h}{2} }{ {9h}^{2}}$

$\displaystyle\sf =\dfrac{1}{9} \:\lim_{h\to 0}\dfrac{sin\dfrac{h}{2} }{h} \times \:\lim_{h\to 0}\dfrac{sin\dfrac{h}{2} }{h}$

$\displaystyle\sf =\dfrac{1}{9} \:\lim_{h\to 0}\dfrac{sin\dfrac{h}{2} }{\dfrac{h}{2} \times 2} \times \:\lim_{h\to 0}\dfrac{sin\dfrac{h}{2} }{\dfrac{h}{2} \times 2 }$

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

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

Additional Information :-

$\begin{gathered}(1)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{tan \: x}{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(2)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{log(1 \: + \: x)}{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(3)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{ {e}^{x} - 1 }{x} \: = \: 1 }}}}}} \\ \end{gathered}$

$\begin{gathered}(4)\:{\underline{\boxed{\bf{\blue{{\tt \:\lim_{x\to 0} \: \dfrac{ {a}^{x} - 1 }{x} \: = \: log(a) }}}}}} \\ \end{gathered}$

Answer 2

EXPLANATION.

$\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[\dfrac{2 - \sqrt{3}cosx - sinx }{(6x - \pi)^{2} } \bigg].$

As we know that,

First we put the value of x = π/6 in equation and check their indeterminant form, we get.

$\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[\dfrac{2 - \sqrt{3} cos \times \dfrac{\pi}{6} - sin \times \dfrac{\pi}{6} }{(6 \times \dfrac{\pi}{6} - \pi)^{2} } \bigg].$

$\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[ \dfrac{2 - \sqrt{3} \times cos30 \degree - sin 30 \degree }{(\pi - \pi)^{2} } \bigg]$

$\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[\dfrac{2 - \sqrt{3} \times \dfrac{\sqrt{3} }{2} - \dfrac{1}{2} }{(\pi - \pi)^{2} } \bigg].$

$\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[ \dfrac{2 - \dfrac{3}{2} - \dfrac{1}{2} }{(\pi - \pi)^{2} } \bigg] = \dfrac{0}{0}$

As we can see that,

It is in the form of 0/0 indeterminant.

Now, we can apply L-HOSPITAL'S RULE, we get.

$\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[\dfrac{\dfrac{d(2)}{dx} - \dfrac{d(\sqrt{3} cosx)}{dx} - \dfrac{d(sinx)}{dx} }{\bigg(\dfrac{d(6x)}{dx} - \dfrac{d(\pi)}{dx} \bigg)^{2} } \bigg].$

$\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[\dfrac{\sqrt{3} sinx - cosx}{2 (6x - \pi) \times 6} \bigg]$

$\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[ \dfrac{\sqrt{3} sinx - cosx }{12(6x - \pi)} \bigg].$

Now, we can again apply L-HOSPITAL'S RULE, we get.

$\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[\dfrac{\sqrt{3}cosx + sinx }{72} \bigg].$

Put the value of x = π/6 in equation, we get.

$\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[ \dfrac{\sqrt{3} \times cos \times \dfrac{\pi}{6} + sin \times \dfrac{\pi}{6} }{72} \bigg].$

$\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[ \dfrac{\sqrt{3} \times \dfrac{\sqrt{3} }{2} + \dfrac{1}{2} }{72} \bigg].$

$\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[\dfrac{\dfrac{3}{2} + \dfrac{1}{2} }{72} \bigg]. = \dfrac{1}{36}$

$\sf \implies \displaystyle \lim_{x \to \frac{\pi}{6} } \bigg[\dfrac{2 - \sqrt{3}cosx - sinx }{(6x - \pi)^{2} } \bigg]. = \dfrac{1}{36}$

                                                                                                                             

MORE INFORMATION.

$\sf \implies (1)= \displaystyle \lim_{x \to \infty} \bigg( 1 + \dfrac{a}{x} \bigg)^{x} = \lim_{x \to 0} ( 1 + ax)^{1/x} = e^{a}$

$\sf \implies(2) = \displaystyle \lim_{x \to 0} \dfrac{a^{x} - 1}{x} = log_{e}a (a>0)$

$\sf \implies (3) =\displaystyle \lim_{x \to0} = \dfrac{e^{x} - 1}{x} = 1$

$\sf \implies(4) = \displaystyle \lim_{x \to a} \dfrac{x^{n} - a^{n} }{x - a} = n a^{n - 1}$

$\sf \implies (5) = \displaystyle \lim_{x \to 0} \dfrac{log(1 + x)}{x} = 1.$

$\sf \implies (6) = \displaystyle \lim_{x \to 0} \dfrac{(1 + x)^{n} - 1}{x} = n$

$\sf \implies(7) = \displaystyle \lim_{x \to \infty} \dfrac{sinx}{x} = \lim_{n \to \infty} \dfrac{cosx}{x} = 0.$

$\sf \implies(8) = \displaystyle \lim_{x \to \infty} \dfrac{sin(1/x)}{(1/x)} = 1.$