$\lim_{x\to\infty} (x+\log x)\cdot \log\biggl(\frac{x^2+3x}{x^2-5x}\biggr)$

Question

The limit

$$\lim_{x\to\infty} (x+\log x)\cdot \log\biggl(\frac{x^2+3x}{x^2-5x}\biggr)$$

can be easily calculated by applying L'Hopital rule and its value is $8$. But is there a quicker way to calculate it without applying L'Hopital rule?

Answer 1

So, we know that: $$\log(1+f(x)) \;\; \sim \;\; f(x) \;\; \text{when}\;\; f(x) \to 0$$

Thus, we can transform the limit into the form: $$\lim_{x\to\infty} (x+\log x)\cdot \log\biggl(\frac{x^2-8x+3x+8x}{x^2-5x}\biggr)=\lim_{x\to\infty} (x+\log x)\cdot \log\biggl(1+\frac{8x}{x^2-5x}\biggr)\;\; \sim\;\;\lim_{x\to\infty} x\cdot \log\biggl(1+\frac{8x}{x^2}\biggr)\;\;\sim\;\; \lim_{x\to\infty} x\cdot \frac{8}{x}=8$$

Answer 2

Hint.

Assuming $\log x = \ln x$, this limit is equivalent to

$$ \lim_{x\to\infty}\ln\left[\left(\frac{1+\frac 3x}{1-\frac 5x}\right)^x\left(\frac{1+\frac 3x}{1-\frac 5x}\right)^{\ln x}\right] = \ln\left[\lim_{x\to\infty}\left(\frac{1+\frac 3x}{1-\frac 5x}\right)^x\lim_{x\to\infty}\left(\frac{1+\frac 3x}{1-\frac 5x}\right)^{\ln x}\right] $$

NOTE

$$ \frac{1+\frac 3x}{1-\frac 5x} = \frac{1+\frac 8x-\frac 5x}{1-\frac 5x} = 1+\frac{8}{x\left(1-\frac 5x\right)}\approx 1+\frac 8x $$