Question

Assertion \((\mathbf{A}):\) Let \(\mathrm{f}:(0, \infty) \rightarrow \mathrm{R}\) be a twice continuously differentiable function such that $\left|f^{\prime}(x)+2 x f^{\prime}(x)+\left(x^{2}+1\right) f(x)\right| \leq 1\( for all \)x$ Then \(\lim _{x \rightarrow \infty} f(x)=0\). Reason (R) : Applying L'Hospital's rule twice on the function $\frac{f(x) e^{\frac{x^{2}}{2}}}{e^{\frac{x^{3}}{2}}}\( we get \)\lim _{x \rightarrow \infty} f(x)=0$.

Short answer

Question: Show that the limit of the given function f(x) as x tends to infinity is zero, given that the function satisfies the inequality: $$\left|f^{\prime\prime}(x)+2 x f^{\prime}(x)+\left(x^{2}+1\right) f(x)\right| \leq 1$$ Answer: By applying L'Hospital's rule twice on the function $$\frac{f(x)e^{\frac{x^{2}}{2}}}{e^{\frac{x^{3}}{2}}}$$ and using the given inequality, we have shown that $$\lim _{x \rightarrow \infty} f(x) = 0$$

Step-by-step solution

Verifying the conditions for L'Hospital's rule

We are given a function in Reason (R) to apply L'Hospital's rule on: $$\frac{f(x)e^{\frac{x^{2}}{2}}}{e^{\frac{x^{3}}{2}}}$$ Denote this function as g(x). We want to find lim(x→∞)g(x). First, let us check the conditions for L'Hospital's Rule: Both the numerator and the denominator should tend to 0 or ∞ as x→∞. On inspection, we clearly see that both the numerator and the denominator tend to infinity as x→∞. Hence, we can apply L'Hospital's rule on g(x).

Applying L'Hospital's rule for the first time

Let us differentiate the numerator and denominator of g(x) with respect to x, and then divide the results: $$\frac{d}{dx}\left(f(x)e^{\frac{x^{2}}{2}}\right) = e^{\frac{x^{2}}{2}}(f^{\prime}(x) + xf(x))$$ and $$\frac{d}{dx}\left(e^{\frac{x^{3}}{2}}\right) =\frac{3x^{2}}{2}e^{\frac{x^{3}}{2}}$$ So after the first application of L'Hospital's rule, we have: $$g'(x) = \frac{e^{\frac{x^{2}}{2}}(f^{\prime}(x) + xf(x))}{\frac{3x^{2}}{2}e^{\frac{x^{3}}{2}}}$$ Now we proceed to apply L'Hospital's rule for the second time.

Applying L'Hospital's rule for the second time

Let us differentiate the numerator and the denominator of g'(x) with respect to x, and then divide the results: $$\frac{d}{dx}\left(e^{\frac{x^{2}}{2}}(f^{\prime}(x) + xf(x))\right) = e^{\frac{x^{2}}{2}}(f^{\prime\prime}(x) + 2xf^{\prime}(x) + (x^{2}+1)f(x))$$ and $$\frac{d}{dx}\left(\frac{3x^{2}}{2}e^{\frac{x^{3}}{2}}\right) = \frac{3}{2}(x^{2}+2x^{3})e^{\frac{x^{3}}{2}}$$ So after the second application of L'Hospital's rule, we have: $$g''(x) = \frac{e^{\frac{x^{2}}{2}}(f^{\prime\prime}(x) + 2xf^{\prime}(x) + (x^{2}+1)f(x))}{\frac{3}{2}(x^{2}+2x^{3})e^{\frac{x^{3}}{2}}}$$

Checking the limit as x approaches infinity

Notice that we can now use the given inequality of the function f(x) for the numerator of g''(x): $$\left|f^{\prime\prime}(x) + 2xf^{\prime}(x) + (x^{2}+1)f(x)\right| \leq 1$$ This implies that $$-\frac{1}{e^{\frac{x^{2}}{2}}} \leq g''(x) \leq \frac{1}{e^{\frac{x^{2}}{2}}}$$ Now, as x approaches infinity, both the left and right hand sides of the inequality converge to 0. Using the squeeze theorem, we conclude that $$\lim _{x \rightarrow \infty} g''(x)=0$$ Since L'Hospital's rule was applied twice, we have $$\lim _{x \rightarrow \infty} g(x) = \lim _{x \rightarrow \infty} \frac{f(x)e^{\frac{x^{2}}{2}}}{e^{\frac{x^{3}}{2}}} = \lim _{x \rightarrow \infty} f(x) = 0$$ Thus, using L'Hospital's rule twice on the given function proved the assertion that \(\lim _{x \rightarrow \infty} f(x)=0\).