## Problem Statement

Calculate the limit of the function:

$\lim _{x \rightarrow-3} \frac{\left(x^{2}+2 x-3\right)^{2}}{x^{3}+4 x^{2}+3 x}$
## Solution

$\lim _{x \rightarrow-3} \frac{\left(x^{2}+2 x-3\right)^{2}}{x^{3}+4 x^{2}+3 x}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-3} \frac{(x+3)^{2}(x-1)^{2}}{x(x+1)(x+3)}=$

$=\lim _{x \rightarrow-3} \frac{(x+3)(x-1)^{2}}{x(x+1)}=\frac{(-3+3)(-3-1)^{2}}{-3(-3+1)}=\frac{0 \cdot(-4)^{2}}{6}=0$

## Problem Kuznetsov Limits 10-5