Find $\lim_{x \to \infty} \frac{x \ln \left( 1 + \frac{\ln x}{x} \right)}{\ln x}$

Question

$$\lim_{x \rightarrow \infty} \frac{x \ln \left( 1 + \frac{\ln x}{x} \right)}{\ln x}$$ $$\lim_{x \rightarrow \infty} \frac{\ln x}{x} = 0$$ Then the given Question look like this
$$\lim_{x \rightarrow \infty} \frac{x \ln 1}{\ln x} = 0$$

But the answer is 1 .. why this is so..

Answer 1

You noted correctly: $$\lim_{x \rightarrow \infty} \frac{\ln x}{x} = 0,$$ but: $$\lim_{x \rightarrow \infty} \frac{x \ln \left( 1 + \frac{\ln x}{x} \right)}{\ln x}=\lim_{x \rightarrow \infty} \frac{x}{\ln x}\cdot \ln \left( 1 + \frac{\ln x}{x} \right)\ne \underbrace{\lim_{x \rightarrow \infty} \frac{x}{\ln x}}_{=\infty}\cdot\underbrace{\lim_{x \rightarrow \infty}\ln \left( 1 + \frac{\ln x}{x} \right)}_{=0},$$ because the first limit on the right does not exist. See Limit properties.

Alternatively: $$\lim_{x \rightarrow \infty} \frac{x \ln \left( 1 + \frac{\ln x}{x} \right)}{\ln x}=\lim_{x \rightarrow \infty} \ln \left( 1 + \frac{\ln x}{x} \right)^{\frac{x }{\ln x}}=\ln \lim_{x \rightarrow \infty} \left( 1 + \frac{\ln x}{x} \right)^{\frac{x }{\ln x}}=\ln e=1.$$

Answer 2

Let $y = \frac{\ln(x)}{x}$

$$\lim_{x\to\infty} \frac{\ln(x)}{x} = \lim_{x\to\infty} \frac{1/x}{1} = 0 \ ( \text{L'Hopital's rule }) $$

So, as $x\to \infty, y\to 0$

Now, $$l = \lim_{x\to\infty} \frac{x\ln(1+\frac{\ln x}{x})}{\ln x} = \lim_{y\to0}\frac{\ln(1+y)}{y} = 1$$

Answer 3

In general, $$ \lim_{x\to\infty} f(x,g(x))\neq\lim_{x\to\infty}f(x,\lim_{y\to\infty}g(y)) $$ which is why you can't just replace $\frac{\ln x}{x}$ with $0$ into the numerator $x\ln(1+\frac{\ln x}{x})$.

Answer 4

Not an answer, so please don't downvote. Nevertheless, here's a plot: