Limit of $\frac{f^{\prime}(x)}{f(x)}$

Calculus Level 5

Find the value of $\displaystyle \lim \limits_{x \to \infty} \frac{f^{\prime}(x)}{f(x)}$ where $f(x)= \displaystyle \int_{-\frac π4}^{\frac π4} e^{x\tan t} \, dt$ .

0.5

e

1

2

0

\pi

1 solution

Mark Hennings
Mar 7, 2021

The substitutions $u = \tan t$ and $v = 1-u$ give

$\begin{aligned} f(x) & = \; \int_{-1}^1 \frac{e^{ux}}{1 + u^2}\,du \; = \; e^x \int_{-1}^1 \frac{e^{-(1-u)x}}{1+u^2}\,du \; = \; e^x \int_0^2 e^{-xv}g(v)\,dv \; > \; 0 \\ f'(x) & = \; \int_{-1}^1 \frac{ue^{ux}}{1 + u^2}\,du \; = \; e^x\int_0^2 e^{-xv}h(v)\,dv\end{aligned}$ where $g(v) \; = \; \frac{1}{1 + (1-v)^2} \hspace{2cm} h(v) \; = \; \frac{1-v}{1 + (1-v)^2}$ Using Watson's Lemma, we deduce that $e^{-x}f(x) \; \sim \; g(0)\Gamma(1)x^{-1} \; = \; \frac{1}{2x} \hspace{1cm} e^{-x}f'(x) \; \sim \; h(0)\Gamma(1)x^{-1} \; =\; \frac{1} {2x} \hspace{2cm} x \to \infty$ and hence

$\lim_{x \to \infty} \frac{f'(x)}{f(x)} \; =\; \boxed{1}$

Whoa! Thanks for the solution! I’m curious tho if this problem can be solved in an easier way...

Inquisitor Math - 3 months, 1 week ago

Well, it is not difficult to obtain Watson's Lemma in this special case. Since $g$ is continuously differentiable on $[0,2]$ , we can find $K > 0$ such that $|g(u) - g(0)| \le Ku$ for $0 \le u \le 2$ . Thus $\begin{aligned} \left| e^{-x}f(x) - \frac{g(0)}{x}(1 - e^{-2x})\right| & =\; \left|\int_0^2 (g(u) - g(0))e^{-xu}\,du\right| \; \le \; \int_0^2 |g(u) - g(0)|e^{-xu}\,du \\ &\le \; K\int_0^2 u e^{-xu}\,du \; = \; K\left(-\frac{2}{x}e^{-2x} + \frac{1}{x}\int_0^2 e^{-xu}\,du\right) \; \le \; K\left(-\frac{2}{x}e^{-2x} + \frac{1}{x^2}(1 - e^{-2x})\right) \end{aligned}$ and so certainly $\lim_{x \to \infty} xe^{-x}f(x) = g(0)$ . SImilarly for $f'(x)$ .

Mark Hennings - 3 months, 1 week ago

Limit of f ′ ( x ) f ( x ) \frac{f^{\prime}(x)}{f(x)} f ( x ) f ′ ( x ) ​

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Limit of $\frac{f^{\prime}(x)}{f(x)}$