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Limit of f at infinity when f'' + 2xf' + (x^2 + 1)f is bounded | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question
Let f:(0,∞)→Rf:\left(0,\infty\right) \to \mathbb R be a twice differentiable function such that
∣f′′(x)+2xf′(x)+(x2+1)f(x)∣≤2∀ x∈R,\left|f''\left(x\right)+2xf'\left(x\right)+\left(x^2+1\right)f\left(x\right)\right| \le 2 \qquad \forall\,x \in \mathbb R ,
 and lim⁡x→∞(xf(x)+f′(x))=5\displaystyle\lim_{x\to\infty}\left(xf\left(x\right)+f'\left(x\right)\right) = 5. Then lim⁡x→∞f(x)\displaystyle\lim_{x\to\infty}f\left(x\right) is equal to
Solution
Answer: 0
Step 1: Let E(x)=e x2/2E\left(x\right) = e^{\,x^2/2}.
ddx[f(x)E(x)]=[f′(x)+xf(x)]E(x)\frac{d}{dx}\left[f\left(x\right)E\left(x\right)\right] = \left[f'\left(x\right)+xf\left(x\right)\right]E\left(x\right)
d2dx2[f(x)E(x)]=[f′′+xf′+f+x(f′+xf)]E=[f′′(x)+2xf′(x)+(x2+1)f(x)]E(x)\frac{d^2}{dx^2}\left[f\left(x\right)E\left(x\right)\right] = \left[f''+xf'+f+x\left(f'+xf\right)\right]E = \left[f''\left(x\right)+2xf'\left(x\right)+\left(x^2+1\right)f\left(x\right)\right]E\left(x\right)
 So the bounded quantity in the hypothesis is exactly [fE]′′\left[fE\right]'' divided by EE. Step 2:
ddxE(x)=xE(x),d2dx2E(x)=(1+x2)E(x)\frac{d}{dx}E\left(x\right) = xE\left(x\right),\qquad \frac{d^2}{dx^2}E\left(x\right) = \left(1+x^2\right)E\left(x\right)
 Step 3: Writing ff as a quotient,
lim⁡x→∞f(x)=lim⁡x→∞f(x)E(x)E(x)\lim_{x\to\infty}f\left(x\right) = \lim_{x\to\infty}\frac{f\left(x\right)E\left(x\right)}{E\left(x\right)}
 The denominator E(x)→∞E\left(x\right) \to \infty, so the ∗∞\dfrac{\ast}{\infty} form of L'Hopital applies:
=lim⁡x→∞[f′(x)+xf(x)]E(x)xE(x)=lim⁡x→∞f′(x)+xf(x)x=0= \lim_{x\to\infty}\frac{\left[f'\left(x\right)+xf\left(x\right)\right]E\left(x\right)}{xE\left(x\right)} = \lim_{x\to\infty}\frac{f'\left(x\right)+xf\left(x\right)}{x} = 0
 the second hypothesis making the numerator tend to the finite value 55 while x→∞x \to \infty. Step 4: Applying L'Hopital once more to the quotient of Step 3, or twice from the start,
lim⁡x→∞f(x)=lim⁡x→∞[f′′+2xf′+(x2+1)f]E(1+x2)E=lim⁡x→∞f′′(x)+2xf′(x)+(x2+1)f(x)1+x2\lim_{x\to\infty}f\left(x\right) = \lim_{x\to\infty}\frac{\left[f''+2xf'+\left(x^2+1\right)f\right]E}{\left(1+x^2\right)E} = \lim_{x\to\infty}\frac{f''\left(x\right)+2xf'\left(x\right)+\left(x^2+1\right)f\left(x\right)}{1+x^2}
 The numerator has absolute value at most 22 by hypothesis, and the denominator →∞\to\infty, so by the squeeze theorem
∣lim⁡x→∞f(x)∣≤lim⁡x→∞21+x2=0\left|\lim_{x\to\infty}f\left(x\right)\right| \le \lim_{x\to\infty}\frac2{1+x^2} = 0
lim⁡x→∞f(x)=0\lim_{x\to\infty}f\left(x\right) = 0
 Answer: 00
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