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

Question

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).

Answer 1

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

Answer 2

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

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