## PROBLEM 3

a) Determine $m \in \mathbb{R}$, such that $\lim _{x \rightarrow-\infty} \frac{\sqrt{(m-1)^{2} \cdot x^{2}+1}}{3 x+2}=-1$.

b) Calculate the limit: $\lim _{x \rightarrow 0} \frac{\ln \left(x^{n}+x^{n-1}+\ldots+x+1\right)}{n x}, n \in \mathbb{N}^{*}$.
Solution:
a)

$\lim _{x \rightarrow-\infty} \frac{\sqrt{(m-1)^{2} \cdot x^{2}+1}}{3 x+2}=\lim _{x \rightarrow-\infty} \frac{|x| \cdot \sqrt{(m-1)^{2}+\frac{1}{x^{2}}}}{3 x+2}=\lim _{x \rightarrow-\infty} \frac{-x \cdot \sqrt{(m-1)^{2}+\frac{1}{x^{2}}}}{x\left(3+\frac{2}{x}\right)}=\frac{-|m-1|}{3}$

2 points

We obtain: $\frac{-|m-1|}{3}=-1 \Rightarrow|m-1|=3$, from which $m \in\{-2,4\}$
b) The limit can be written in the form:

$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\ln \left[1+\left(x+x^{2}+\ldots+x^{n}\right)\right]}{x+x^{2}+\ldots+x^{n}} \cdot \frac{x+x^{2}+\ldots+x^{n}}{n x}=1 \cdot \lim _{x \rightarrow 0} \frac{x+x^{2}+\ldots+x^{n}}{n x}= \\
& =\lim _{x \rightarrow 0} \frac{x\left(1+x+x^{2}+\ldots+x^{n-1}\right)}{n x}=\lim _{x \rightarrow 0} \frac{1+x+x^{2}+\ldots+x^{n-1}}{n}=\frac{1}{n}
\end{aligned}
$$

4 points

Note: $\lim _{x \rightarrow \alpha} \frac{\ln [1+u(x)]}{u(x)}=1$ was used, for $u(x) \rightarrow 0$.