Answer
$+ \infty$
Work Step by Step
Our aim is to compute the value of $\lim\limits_{x \to \ -1^{+}}\dfrac{x^2+1}{x^2+3x+2}$,
Here, we need to use the formula of:
$\lim \limits_{x\to a}f(x)=f(a)$, this implies that:
$\lim\limits_{x \to \ -1^{+}}\dfrac{x^2+1}{x^2+3x+2}=\dfrac{(-0.99)^2+1}{(-0.99)^2+3(-0.99)+2}\\=+ \infty$
