Question

Find out the answer syep by step

::☆☆Moderators

::☆☆Stars

only or the best answers

NO SPAM
$\displaystyle \lim_{x \to \frac{\pi}{4} } \frac{ {cot}^{3} x - tanx}{cos( \frac{\pi}{4} + x)}$
​

Answer 1

Solution

$\displaystyle \lim_{x \to \frac{\pi}{4} } \frac{ {cot}^{3} x - tanx}{cos( \frac{\pi}{4} + x)} \\ \\ = \displaystyle \lim_{x \to \frac{\pi}{4} } \frac{ 1 - {tan}^{4}x }{{ \dfrac{1}{ \sqrt{2} } (cosx - sinx )} } \times \frac{1}{ {tan}^{3}x } \\ \\ = \displaystyle \lim_{x \to \frac{\pi}{4} } \frac{(1 - {tan}^{2}x) }{cosx - sinx} \times \frac{ \sqrt{2}(1 + {tan}^{2} x) }{ {tan}^{3} x} \\ \\ = \displaystyle \lim_{x \to \frac{\pi}{4} } \frac{(1 - {tan}^{2}x) }{cosx - sinx} \times \frac{ \sqrt{2}({sec}^{2} x) }{ {tan}^{3} x} \\ \\ = \displaystyle \lim_{x \to \frac{\pi}{4} } \frac{ {cos}^{2}x - {sin}^{2} x }{cosx - sinx} \times \frac{ \sqrt{2}( {sec}^{2} x) }{ cos {}^{2} x. {tan}^{3x} } \\ \\ \displaystyle \lim_{x \to \frac{\pi}{4} } \frac{(cosx + sinx) \times \cancel{(cosx - sinx)}}{\cancel{(cosx - sinx)}} \times \frac{ \sqrt{2}( {sec}^{4} x) }{ {tan}^{3x} } \\ \\ = \displaystyle \lim_{x \to \frac{\pi}{4} } \frac{ \sqrt{2} {sec} {}^{4}x }{ {tan}^{3}x } (cosx + sinx)$

On applying Limit

$= \dfrac{ \sqrt{2} {( \sqrt{2} )}^{4} }{1^{3}}( \dfrac{1}{ \sqrt{2} } + \dfrac{1}{ \sqrt{2} } ) \\ \\ = 4 \sqrt{2} ( \frac{2}{ \sqrt{2} } ) \\ \\ = 8$

Which is the required answer

Answer 2

Question :

$\displaystyle \dagger\ \; \sf \red{\lim_{x \to \frac{\pi}{4}} \dfrac{cot^3x-tan\ x}{cos\left( \frac{\pi}{4} + x \right)}}$

Solution :

$\displaystyle \sf \lim_{x \to \frac{\pi}{4}} \dfrac{cot^3x-tan\ x}{cos\left( \frac{\pi}{4} + x \right)}$

On plugging the values directly into the eq. , it leads to indeterminant form  $\dfrac{0}{0}$

$\bullet\ \; \sf \orange{cot\ x = \dfrac{1}{tan\ x}}$

$\bullet\ \; \sf \green{cos(A+B)=cos\ A\ .cos\ B-sin\ A\ .sin\ B}$

$\longrightarrow \displaystyle \sf \lim_{x \to \frac{\pi}{4}} \dfrac{ \frac{1}{tan^3x} -tan\ x}{cos\ \frac{\pi}{4}.cos\ x- sin\ \frac{\pi}{4}.sin\ x}$

$\bullet\ \; \sf \blue{cos\ \dfrac{\pi}{4}=sin\ \dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}}$

$\longrightarrow \displaystyle \sf \lim_{x \to \frac{\pi}{4}} \dfrac{ \frac{1-tan^4x}{tan^3x}}{\frac{1}{\sqrt{2}}\left(cos\ x-sin\ x\right)}$

$\longrightarrow \displaystyle \sf \lim_{x \to \frac{\pi}{4}} \dfrac{ 1-tan^4x}{(cos\ x-sin\ x)} \times \dfrac{\sqrt{2}}{tan^3x}$

$\bullet\ \; \sf \orange{1-A^4=1^2-(A^2)^2=(1-A^2)(1+A^2)}$

$\longrightarrow \displaystyle \sf \lim_{x \to \frac{\pi}{4}} \dfrac{ 1-tan^2x}{(cos\ x-sin\ x)} \times \dfrac{\sqrt{2}(1+tan^2x)}{tan^3x}$

$\bullet\ \; \sf \green{sec^2x=1+tan^2x}$

$\bullet\ \; \sf \blue{tan\ x=\dfrac{sin\ x}{cos\ x}}$

$\longrightarrow \displaystyle \sf \lim_{x \to \frac{\pi}{4}} \dfrac{ 1-\frac{sin^2x}{cos^2x}}{(cos\ x-sin\ x)} \times \dfrac{\sqrt{2}(sec^2x)}{tan^3x}$

$\longrightarrow \displaystyle \sf \lim_{x \to \frac{\pi}{4}} \dfrac{ cos^2x-sin^2x}{(cos\ x-sin\ x)} \times \dfrac{\sqrt{2}(sec^2x)}{cos^2x.tan^3x}$

$\bullet\ \; \sf \orange{sec^2x=\dfrac{1}{cos^2x}}\\\\\bullet\ \; \sf \green{cos^2x=\dfrac{1}{sec^2x}}$

$\\\longrightarrow \displaystyle \sf \lim_{x \to \frac{\pi}{4}} \dfrac{ (cos\ x-sin\ x)(cos\ x+sin\ x)}{(cos\ x-sin\ x)} \times \dfrac{\sqrt{2}(sec^4x)}{tan^3x}$

$\longrightarrow \displaystyle \sf \lim_{x \to \frac{\pi}{4}} (cos\ x+sin\ x )\left(\dfrac{\sqrt{2}(sec^4x)}{tan^3x}\right)$

Sub. the limit value ,

$\longrightarrow \sf \left( cos\ \dfrac{\pi}{4}+sin\ \dfrac{\pi}{4} \right) \left(\dfrac{\sqrt{2}(sec^4 \frac{\pi}{4} )}{tan^3 \frac{\pi}{4} }\right)$

$\bullet\ \; \sf \blue{cos\ \dfrac{\pi}{4}=sin\ \dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}}$

$\bullet\ \; \sf \orange{sec\ \dfrac{\pi}{4}= \sqrt{2}}$

$\bullet\ \; \sf \green{tan\ \dfrac{\pi}{4}=1}$

$\longrightarrow \sf \left( \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}} \right) \left(\dfrac{\sqrt{2}(\sqrt{2})^4}{(1)^3 }\right)$

$\longrightarrow \sf \dfrac{2}{\sqrt{2}} \times 4\sqrt{2}$

$\longrightarrow\ \; \sf \pink{8}$

★ ═════════════════════ ★

$\bullet\ \; \displaystyle \bf \purple{\lim_{x \to \frac{\pi}{4}} \dfrac{cot^3x-tan\ x}{cos\left( \frac{\pi}{4} + x \right)} =8}$