Question

question 4
if (1leq f(x)leq x^{2}+7x + 7), for all (x), find (lim_{x
ightarrow - 1}f(x)).
( -\frac{1}{16})
1
8
does not exist
(-\frac{1}{8})

Explanation:

Step1: Find left - hand limit of lower - bound

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

Step2: Find left - hand limit of upper - bound

Calculate $\lim_{x
ightarrow - 1}(x^{2}+7x + 7)$. Substitute $x=-1$ into $x^{2}+7x + 7$. We get $(-1)^{2}+7\times(-1)+7=1 - 7 + 7=1$.

Step3: Apply Squeeze Theorem

Since $1\leq f(x)\leq x^{2}+7x + 7$ for all $x$ and $\lim_{x
ightarrow - 1}1=\lim_{x
ightarrow - 1}(x^{2}+7x + 7)=1$, by the Squeeze Theorem, $\lim_{x
ightarrow - 1}f(x)=1$.

Answer:

1