JEE Main 2022 — Mathematics Calculus

Given,$\underset{x\rightarrow 0}{lim}{(\frac{{(x+2cosx)}^{3}+2{(x+2cosx)}^{2}+3sin(x+2cosx)}{{(x+2)}^{3}+2{(x+2)}^{2}+3sin(x+2)})}^{\frac{100}{x}}$x→0lim​((x+2)3+2(x+2)2+3sin(x+2)(x+2cosx)3+2(x+2cosx)2+3sin(x+2cosx)​)x100​

Now by putting the value of limit we can see it is in form ${1}^{\infty }$1∞

Now we know that,

$\underset{x\rightarrow 0}{\mathrm{lim}}{(f(x))}^{g(x)}$x→0lim​(f(x))g(x) when is in form of ${1}^{\infty }$1∞, we can write this as ${e}^{\underset{x\rightarrow 0}{\mathrm{lim}}(f(x)-1)g(x)}$ex→0lim​(f(x)−1)g(x)

Now using the above rule we get,

$$={e}^{\underset{x\rightarrow 0}{\mathrm{lim}}[(\frac{{(x+2\mathrm{cos}x)}^{3}+2{(x+2\mathrm{cos}x)}^{2}+3\mathrm{sin}(x+2\mathrm{cos}x)}{{(x+2)}^{3}+2{(x+2)}^{2}+3\mathrm{sin}(x+2)})-1]\times \frac{100}{x}}$$=ex→0lim​[((x+2)3+2(x+2)2+3sin(x+2)(x+2cosx)3+2(x+2cosx)2+3sin(x+2cosx)​)−1]×x100​

$$={e}^{\underset{x\rightarrow 0}{\mathrm{lim}}[\frac{100}{x}[\frac{{(x+2\mathrm{cos}x)}^{3}+2{(x+2\mathrm{cos}x)}^{2}+3\mathrm{sin}(x+2\mathrm{cos}x)-({(x+2)}^{3}+2{(x+2)}^{2}+3\mathrm{sin}(x+2))}{{(x+2)}^{3}+2{(x+2)}^{2}+3\mathrm{sin}(x+2)}]]}$$=ex→0lim​[x100​[(x+2)3+2(x+2)2+3sin(x+2)(x+2cosx)3+2(x+2cosx)2+3sin(x+2cosx)−((x+2)3+2(x+2)2+3sin(x+2))​]]

$$={e}^{\underset{x\rightarrow 0}{\mathrm{lim}}\frac{100}{x}[(\frac{{(x+2\mathrm{cos}x)}^{3}-{(x+2)}^{3}+2{(x+2\mathrm{cos}x)}^{2}-2{(x+2)}^{2}+3\mathrm{sin}(x+2\mathrm{cos}x)-3\mathrm{sin}(x+2)}{8+8+3\mathrm{sin}2})]}$$=ex→0lim​x100​[(8+8+3sin2(x+2cosx)3−(x+2)3+2(x+2cosx)2−2(x+2)2+3sin(x+2cosx)−3sin(x+2)​)]

Using L-hospital method we get,

$$={e}^{\frac{100}{16+3\mathrm{sin}2}\underset{x\rightarrow 0}{\mathrm{lim}}\frac{3{(x+2\mathrm{cos}x)}^{2}\times (1+2\mathrm{sin}x)-3{(x+2)}^{2}+4(x+2\mathrm{cos}x)\times (1-2\mathrm{sin}x)-4(x+2)+3\mathrm{cos}(x+2\mathrm{cos}x)\times (1-2\mathrm{sin}x)-3\mathrm{cos}(x+2)}{1}}$$=e16+3sin2100​x→0lim​13(x+2cosx)2×(1+2sinx)−3(x+2)2+4(x+2cosx)×(1−2sinx)−4(x+2)+3cos(x+2cosx)×(1−2sinx)−3cos(x+2)​

$$={e}^{\frac{100}{16+3\mathrm{sin}2}(\frac{12-3(4)+8\times 1-8+3\mathrm{cos}2-3\mathrm{cos}2}{1})}$$=e16+3sin2100​(112−3(4)+8×1−8+3cos2−3cos2​)

$$={e}^{\frac{100}{16+3\mathrm{sin}2}\times 0}$$=e16+3sin2100​×0

$$={e}^{0}=1$$=e0=1