JEE Main 2019 — Mathematics Calculus

$\underset{x\rightarrow \frac{\pi }{4}}{lim}\frac{{cot}^{3}x-tanx}{cos(x+\frac{\pi }{4})}$x→4π​lim​cos(x+4π​)cot3x−tanx​$=\underset{x\rightarrow \frac{\pi }{4}}{\mathrm{lim}}\frac{(\frac{1}{{\mathrm{tan}}^{3}x}-\mathrm{tan}x)}{(\frac{1}{\sqrt{2}}\mathrm{cos}x-\frac{1}{\sqrt{2}}\mathrm{sin}x)}[\because \mathrm{cot}\theta =\frac{1}{\mathrm{tan}\theta }]$=x→4π​lim​(2​1​cosx−2​1​sinx)(tan3x1​−tanx)​[∵cotθ=tanθ1​]

$$=\underset{x\rightarrow \frac{\pi }{4}}{lim}\frac{(1-ta{n}^{4}x)}{ta{n}^{3}x.(\frac{1}{\sqrt{2}}cosx-\frac{1}{\sqrt{2}}sinx)}$$=x→4π​lim​tan3x.(2​1​cosx−2​1​sinx)(1−tan4x)​

$$=\underset{x\rightarrow \frac{\pi }{4}}{\mathrm{lim}}\frac{(1-{\mathrm{tan}}^{2}x)(1+{\mathrm{tan}}^{2}x)}{{\mathrm{tan}}^{3}x.(\frac{1}{\sqrt{2}}\mathrm{cos}x-\frac{1}{\sqrt{2}}\mathrm{sin}x)}$$=x→4π​lim​tan3x.(2​1​cosx−2​1​sinx)(1−tan2x)(1+tan2x)​

=\underset{x\rightarrow \frac{\pi }{4}}{lim}\frac{\sqrt{2}[(1-\mathrm{tan}x)(1+\mathrm{tan}x){\mathrm{sec}}^{2}x]}{ta{n}^{3}x.(cosx-sinx)}[\because 1+{\mathrm{tan}}^{2}x={\mathrm{sec}}^{2}x&{a}^{2}-{b}^{2}=(a-b)(a+b)]

$$=\underset{x\rightarrow \frac{\pi }{4}}{\mathrm{lim}}\frac{\sqrt{2}[(1-\frac{\mathrm{sin}x}{\mathrm{cos}x})(1+\mathrm{tan}x){\mathrm{sec}}^{2}x]}{{\mathrm{tan}}^{3}x.(\mathrm{cos}x-\mathrm{sin}x)}$$=x→4π​lim​tan3x.(cosx−sinx)2​[(1−cosxsinx​)(1+tanx)sec2x]​

$$=\underset{x\rightarrow \frac{\pi }{4}}{lim}\frac{\sqrt{2}(\mathrm{cos}x-\mathrm{sin}x).(1+\mathrm{tan}x).se{c}^{2}x}{cosx.ta{n}^{3}x.(\mathrm{cos}x-\mathrm{sin}x)}$$=x→4π​lim​cosx.tan3x.(cosx−sinx)2​(cosx−sinx).(1+tanx).sec2x​

$$=\underset{x\rightarrow \frac{\pi }{4}}{\mathrm{lim}}\frac{\sqrt{2}(1+\mathrm{tan}x)}{\mathrm{cos}x.\frac{{\mathrm{sin}}^{3}x}{{\mathrm{cos}}^{3}x}.{\mathrm{cos}}^{2}x}$$=x→4π​lim​cosx.cos3xsin3x​.cos2x2​(1+tanx)​

$$=\underset{x\rightarrow \frac{\pi }{4}}{\mathrm{lim}}\frac{\sqrt{2}(1+\mathrm{tan}x)}{{\mathrm{sin}}^{3}x}$$=x→4π​lim​sin3x2​(1+tanx)​

$$=\frac{\sqrt{2}(1+\mathrm{tan}\frac{\pi }{4})}{{\mathrm{sin}}^{3}\frac{\pi }{4}}$$=sin34π​2​(1+tan4π​)​

$=8$=8.