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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).

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    The thing you did with $\sin(x)\approx x$, you can do for the $\ln(\dots)$ as well, though very tedious... – 2017-01-16
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    the result $$e^{\frac{11}{2}}$$ is right – 2017-01-16
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    i know, that's right, but I would like to know if there is another method, different from mine – 2017-01-16
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    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. – 2017-01-16

2 Answers 2

2

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 $1$, $$ \lim_{t\to0^+}\frac{3t+\frac{5}{2}t+o(t)}{t}=3+\frac{5}{2} $$

1

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

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

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    clever, but it is not immediate the first equality – 2017-01-16
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    @M41Npain, I added the details for the first equality. – 2017-01-16
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    Use of $\approx$ symbols makes the whole thing non-rigorous. Calculus methods are exact and not approximate. – 2017-01-17
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    IMO the best answer, as the Taylor requires derivatives anyway and is quasi equivalent to L'Hospital. – 2017-01-17
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    Anyway, there is a flaw: you may not silently discard the term $9/x$ from the limit and then take the $x^{th}$ power. Because that would also allow you to discard $11/x$ and obtain $1^x\to1$. – 2017-01-17
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    @YvesDaoust, but the power is a large number, and $11 +9/x \approx 11$ and $x-5 \approx x$ for $x >>0$. There is no indetermination. – 2017-01-17
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    Yes and $1+11/x\approx1$ and the limit is $1$. Your reasoning is not correct. – 2017-01-17
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    @YvesDaoust, I edited the answer. Take a look if it is fine now! – 2017-01-17
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    @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. – 2017-01-17
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    @YvesDaoust, yeah! This was the idea! :) – 2017-01-17