Question

if (1leq f(x)leq x^{2}+5x + 5), for all (x), find (lim_{x
ightarrow - 1}f(x)).
1
(-\frac{1}{16})
does not exist
(-\frac{1}{8})
8

Explanation:

Step1: Find left - hand limit

We know that $\lim_{x
ightarrow - 1}1 = 1$.

Step2: Find right - hand limit

Calculate $\lim_{x
ightarrow - 1}(x^{2}+5x + 5)$. Substitute $x=-1$ into $x^{2}+5x + 5$:
\[

$$\begin{align*} (-1)^{2}+5\times(-1)+5&=1 - 5 + 5\\ &=1 \end{align*}$$

\]

Step3: Apply Squeeze Theorem

Since $1\leq f(x)\leq x^{2}+5x + 5$ for all $x$ and $\lim_{x
ightarrow - 1}1=\lim_{x
ightarrow - 1}(x^{2}+5x + 5)=1$, by the Squeeze Theorem, $\lim_{x
ightarrow - 1}f(x)=1$.

Answer:

1