9
$\begingroup$

How to evaluate $\lim _{x\to \infty }\left(\frac{x+3}{\sqrt{x^2-5x}}\right)^{x^2\sin\left(1/x\right)}$?

My Try:

$$\lim _{x\to \infty }\left(x^2\sin\left(\frac{1}{x}\right)\ln\left(\frac{x+3}{\sqrt{x^2-5x}}\right)\right) = \lim _{t\to 0 }\left(\frac{1}{t^2}\sin\left(t\right)\ln\left(\frac{\frac{1}{t}+3}{\sqrt{\frac{1}{t^2}-\frac{5}{t}}}\right)\right)$$ Now $\sin(x) \approx x, x \rightarrow 0$ so: $$\approx \lim _{t\to 0 }\left(\frac{1}{t}ln\left(\frac{\left(3t+1\right)\sqrt{-5t+1}}{1-5t}\right)\right)$$

At this point i used the rule of the de l'Hôpital so: $$\lim _{t\to 0 }\left(\frac{1}{t}ln\left(\frac{\left(3t+1\right)\sqrt{-5t+1}}{1-5t}\right)\right) = \lim _{t\to 0}\left(\frac{\frac{-15t+11}{2\left(-5t+1\right)\left(3t+1\right)}}{1}\right) = \frac{11}{2}$$ So: $$\lim _{x\to \infty }\left(\left(\frac{x+3}{\sqrt{x^2-5x}}\right)^{x^2\sin\left(\frac{1}{x}\right)}\right) = \color{red}{e^\frac{11}{2}}$$ Which it is the exact result of the proposed limit.
My question is, there is another method, different from mine to get the same result? (Preferably without resorting to de l'Hôpital rule).

|cite|improve this question
$\endgroup$
  • $\begingroup$ The thing you did with $\sin(x)\approx x$, you can do for the $\ln(\dots)$ as well, though very tedious... $\endgroup$ – Simply Beautiful Art Jan 16 '17 at 13:17
  • $\begingroup$ the result $$e^{\frac{11}{2}}$$ is right $\endgroup$ – Dr. Sonnhard Graubner Jan 16 '17 at 13:19
  • 1
    $\begingroup$ i know, that's right, but I would like to know if there is another method, different from mine $\endgroup$ – Amarildo Jan 16 '17 at 13:21
  • $\begingroup$ I would have done two things differently. I would have expanded the $\ln$ to $\ln(3t+1)-(1/2)\ln(1-5t)$ (maybe you did that, but hid it.) I don't think there's a substantially different way to do the limit. You could use Taylor series but 1. yuck and 2. that's what L'hospital is really doing anyway. $\endgroup$ – B. Goddard Jan 16 '17 at 13:26
2
$\begingroup$

You're making your own life more difficult. ;-) But your idea is good.

After taking the logarithm, apply the substitution $x=1/t$ where it's not restrictive to assume $x>0$ (actually, $x>5$); note that $$ \frac{x+3}{\sqrt{x^2-5x}}=\frac{1+3t}{\sqrt{1-5t}}, $$ so you have $$ \lim_{t\to0^+}\frac{\sin t}{t^2}\ln\frac{1+3t}{\sqrt{1-5t}}= \lim_{t\to0^+}\frac{\ln(1+3t)-\frac{1}{2}\ln(1-5t)}{t} $$ owing to $\lim_{t\to0}\frac{\sin t}{t}=1$ (of course, conditionally to the existence of the last limit).

This can be rewritten $$ 3\lim_{t\to0^+}\frac{\ln(1+3t)}{3t}+ \frac{5}{2}\lim_{t\to0^+}\frac{\ln(1-5t)}{-5t}=3+\frac{5}{2} $$ or with Taylor up to degree $1$, $$ \lim_{t\to0^+}\frac{3t+\frac{5}{2}t+o(t)}{t}=3+\frac{5}{2} $$

|cite|improve this answer
$\endgroup$
1
$\begingroup$

Notice that $$\frac{x+3}{\sqrt{x^2-5x}} = \left(\dfrac{x^2+6x+9}{x^2-5x}\right)^{1/2} = \left(1+ \dfrac{11x+9}{x^2-5x}\right)^{1/2}.$$

By setting $y = \dfrac{x^2-5x}{11x+9}$ and since $x \rightarrow \infty \implies y \rightarrow \infty$, we obtain

$$ x = \dfrac{5+11y + \sqrt{121y^2+146y+25}}{2}.$$

Thus, $$\lim_{x \rightarrow \infty} \left(\dfrac{x+3}{\sqrt{x^2-5x}}\right)^{x/2} = \lim_{y \rightarrow \infty} \left( 1+\dfrac{1}{y}\right)^{\left(\dfrac{5+11y + \sqrt{121y^2+146y+25}}{4}\right)} = $$ $$= \lim_{y \rightarrow \infty} \left( 1+\dfrac{1}{y}\right)^{5/4}\cdot \left( 1+\dfrac{1}{y}\right)^{11y/4}\cdot \left( 1+\dfrac{1}{y}\right)^{11y/4\cdot \left(\sqrt{1+146/(121y)+25/(121y^2)}\right)} = e^{11/2},$$

Since $g(y) = \sqrt{1+ \frac{146}{121y}+\frac{25}{121y^2}}$ is continuous and $\lim_{y \rightarrow \infty} g(y)$ exists.

Yet, $$\lim_{x \rightarrow \infty}\dfrac{\sin(1/x)}{1/x} = 1.$$

Therefore, we have

$$\lim_{x \rightarrow \infty} \left(\dfrac{x+3}{\sqrt{x^2-5x}}\right)^{x^2\sin\left(1/x\right)} = e^{11/2}.$$

|cite|improve this answer
$\endgroup$
  • $\begingroup$ clever, but it is not immediate the first equality $\endgroup$ – Amarildo Jan 16 '17 at 15:29
  • $\begingroup$ @M41Npain, I added the details for the first equality. $\endgroup$ – Alex Silva Jan 16 '17 at 15:41
  • $\begingroup$ Use of $\approx$ symbols makes the whole thing non-rigorous. Calculus methods are exact and not approximate. $\endgroup$ – Paramanand Singh Jan 17 '17 at 5:30
  • 1
    $\begingroup$ IMO the best answer, as the Taylor requires derivatives anyway and is quasi equivalent to L'Hospital. $\endgroup$ – Yves Daoust Jan 17 '17 at 7:16
  • 1
    $\begingroup$ @AlexSilva: splendid effort ! The key was to make the expression $1+1/y$ appear, I guess. You can directly pull the factor $11y/2$ from the expression of $x$ and show that the remaining factor gives exponent $1$, and even include $x\sin(1/x)$ in a single go, but this is just a variant. $\endgroup$ – Yves Daoust Jan 17 '17 at 16:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.