Finding the limit as $n \to \infty $ of $n\ln\left(1+\frac{\ x}{n^2}\right)$

Question

Find

$$\lim_{n\to \infty} n\ln\left(1+\frac{\ x}{n^2}\right)$$

My attempt: $\lim_{n\to \infty} n \left[\ln\left(\frac{\ n^2 +x}{n^2}\right)\right]$

= $\lim_{n\to \infty} n [\ln (n^2 +x) - \ln(n^2)]$

But I'm not sure how to get this out of indeterminate form.

Answer 1

Assume $x>0$. One may use $$ 0<\ln(1+u)<u,\quad u \in (0,1), $$ giving, as $ n \to \infty$, $$ 0<n\ln \left(1+\frac{x}{n^2}\right)<\frac{x}{n} $$ and conclude.

Answer 2

$n\ln\left(1+\frac{\ x}{n^2}\right) = \frac{1}{n}n^2\ln\left(1+\frac{\ x}{n^2}\right) = \frac{1}{n} \ln \left( \left( 1+\frac{x}{n^2} \right)^{n^2} \right)$

If $n\to\infty$, then $ \frac{1}{n} \ln \left( \left( 1+\frac{x}{n^2} \right)^{n^2} \right) \equiv \frac{1}{n} \ln \left( e^x \right) \equiv \frac{x}{n} \equiv 0$