Question

Evaluate the limit without using L'Hôpital's rule:

$\boxed{\lim \limits_{x \to \frac{\pi}{4} } \dfrac{4 \sqrt{2} - ( \cos x + \sin x) ^{5} }{1 - \sin2x}}$​

Answer 1

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

Given expression is

$\rm \: \displaystyle\lim_{x \to \dfrac{\pi}{4} }\rm \dfrac{4 \sqrt{2} - ( \cos x + \sin x) ^{5} }{1 - \sin2x}$

If we substitute directly the value of x, we get

$\rm \: = \: \dfrac{4 \sqrt{2} - {\bigg(cos\dfrac{\pi}{4} + sin\dfrac{\pi}{4}\bigg) }^{5} }{1 - sin\dfrac{\pi}{2}}$

$\rm \: = \: \dfrac{4 \sqrt{2} - {\bigg( \dfrac{1}{ \sqrt{2} } + \dfrac{1}{ \sqrt{2} } \bigg) }^{5} }{1 - 1}$

$\rm \: = \: \dfrac{4 \sqrt{2} - {\bigg( \dfrac{1 + 1}{ \sqrt{2}} \bigg) }^{5} }{0}$

$\rm \: = \: \dfrac{4 \sqrt{2} - {\bigg( \dfrac{2}{ \sqrt{2}} \bigg) }^{5} }{0}$

$\rm \: = \: \dfrac{4 \sqrt{2} - {\bigg( \sqrt{2} \bigg) }^{5} }{0}$

$\rm \: = \: \dfrac{4 \sqrt{2} - 4 \sqrt{2} }{0}$

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

which is indeterminant form.

So, Consider again

$\rm \: \displaystyle\lim_{x \to \dfrac{\pi}{4} }\rm \dfrac{4 \sqrt{2} - ( \cos x + \sin x) ^{5} }{1 - \sin2x}$

We know,

$\rm \: cosx + sinx$

$\rm \: = \: \sqrt{2}\bigg(\dfrac{1}{ \sqrt{2} } sinx + \dfrac{1}{ \sqrt{2} }cosx \bigg)$

$\rm \: = \: \sqrt{2}\bigg(sin\dfrac{\pi}{4}sinx + cos\dfrac{\pi}{4}cosx\bigg)$

$\rm \: = \: \sqrt{2}cos\bigg(x - \dfrac{\pi}{4} \bigg)$

So,

$\rm \: {(cosx + sinx)}^{5} = 4 \sqrt{2} \: {cos}^{5}\bigg(x - \dfrac{\pi}{4} \bigg)$

So, Substitute this value in above expression, we get

$\rm \: = \: \displaystyle\lim_{x \to \dfrac{\pi}{4} }\rm \dfrac{4 \sqrt{2} -4 \sqrt{2} \: cos ^{5}\bigg(x - \dfrac{\pi}{4} \bigg) }{1 - \sin2x}$

$\rm \: = \: \displaystyle\lim_{x \to \dfrac{\pi}{4} }\rm \dfrac{4 \sqrt{2} \bigg[1- \: cos ^{5}\bigg(x - \dfrac{\pi}{4} \bigg)\bigg]}{1 - \sin2x}$

Now, we use method of Substitution, to evaluate this limit.

So, Substitute

$\rm \: x = \dfrac{\pi}{4} - h, \: \: as \: x \to \: \dfrac{\pi}{4} \: \: so \: \: h \to \: 0$

So, we get

$\rm \: = \: 4 \sqrt{2}\displaystyle\lim_{h \to 0}\rm \frac{1 - {cos}^{5}\bigg(\dfrac{\pi}{4} - h - \dfrac{\pi}{4} \bigg) }{1 - sin2\bigg(\dfrac{\pi}{4} - h\bigg) }$

$\rm \: = \: 4 \sqrt{2}\displaystyle\lim_{h \to 0}\rm \frac{1 - {cos}^{5}( - h) }{1 - sin\bigg(\dfrac{\pi}{2} - 2h\bigg) }$

$\rm \: = \: 4 \sqrt{2}\displaystyle\lim_{h \to 0}\rm \frac{1 - {cos}^{5}h}{1 - cos2h }$

We know, from Binomial expansion

$\rm \: 1 - {x}^{5} = (1 - x)(1 + x + {x}^{2} + {x}^{3} + {x}^{4})$

So, using this identity, we get

$\rm \: = 4 \sqrt{2}\displaystyle\lim_{h \to 0}\rm \frac{(1 - cosh)(1 + cosh + {cos}^{2}h + {cos}^{3}h + {cos}^{4}h}{ {2sin}^{2} h}$

$\rm \: = 4 \sqrt{2} \times \displaystyle\lim_{h \to 0}\rm(1 + cosh + {cos}^{2}h + {cos}^{3}h + {cos}^{4}h) \displaystyle\lim_{h \to 0}\rm \frac{2 {sin}^{2} \dfrac{h}{2} }{ {2sin}^{2} h}$

$\rm \: = 4 \sqrt{2} \times (1 + 1 + 1 + 1 + 1) \displaystyle\lim_{h \to 0}\rm \frac{{sin}^{2} \dfrac{h}{2} }{ {sin}^{2} h}$

$\rm \: = 20\sqrt{2} \displaystyle\lim_{h \to 0}\rm \frac{{sin}^{2} \dfrac{h}{2} }{ {\bigg(2 \: sin\dfrac{h}{2} \: cos\dfrac{h}{2} \bigg) }^{2} }$

$\rm \: = 20\sqrt{2} \displaystyle\lim_{h \to 0}\rm \frac{{sin}^{2} \dfrac{h}{2} }{ 4 {sin}^{2}\dfrac{h}{2} \: {cos}^{2}\dfrac{h}{2} }$

$\rm \: = 5\sqrt{2} \displaystyle\lim_{h \to 0}\rm \frac{1 }{ {cos}^{2}\dfrac{h}{2} }$

$\rm \: = \: 5 \sqrt{2} \times 1$

$\rm \: = \: 5 \sqrt{2}$

Hence,

$\: \boxed{\tt{ \rm \: \: \displaystyle\lim_{x \to \dfrac{\pi}{4} }\rm \dfrac{4 \sqrt{2} - ( \cos x + \sin x) ^{5} }{1 - \sin2x} = 5 \sqrt{2} \: \: }}$