JEE Main 2022 (Online) 28th July Morning ShiftQuestion 1
$\lim_{x\to0}\left(\frac{(x+2\cos x)^3+2(x+2\cos x)^2+3\sin(x+2\cos x)}{(x+2)^3+2(x+2)^2+3\sin(x+2)}\right)^{\frac{100}{x}}$ is equal to ________.
Limits & Continuity
Answer$1$
Solution
Concept. $1^\infty$ form. The two arguments differ by order $x^2$, so the exponent tends to $0$. Formulas used. (1) $\lim\Big(\frac{A}{B}\Big)^{k/x}=e^{\lim\frac kx\cdot\frac{A-B}{B}}$ when $\frac AB\to1$(2) $\phi(u)-\phi(v)\approx\phi'(v)(u-v)$ when $u-v\to0$Step 1 - Set up. Let $\phi(t)=t^3+2t^2+3\sin t$, $u=x+2\cos x$, $v=x+2$. The base is $\frac{\phi(u)}{\phi(v)}\to\frac{\phi(2)}{\phi(2)}=1$. Step 2 - Difference. $u-v=2\cos x-2\sim-x^2$. By (2), $\phi(u)-\phi(v)\sim-\phi'(2)x^2$. Step 3 - Exponent. By (1): $$\lim\frac{100}{x}\cdot\frac{-\phi'(2)x^2}{\phi(2)}=0$$ Step 4. Limit $=e^0=1$. Answer: 1