Question

question 4
if 1 ≤ f(x) ≤ x² + 7x + 7, for all x, find lim_{x→ - 1} f(x).
-\frac{1}{16}
1
8
does not exist
-\frac{1}{8}

Explanation:

Step1: Find left - hand limit

The left - hand function is a constant function $y = 1$. So, $\lim_{x
ightarrow - 1}1=1$.

Step2: Find right - hand limit

For the function $y=x^{2}+7x + 7$, substitute $x=-1$ into it. We have $\lim_{x
ightarrow - 1}(x^{2}+7x + 7)=(-1)^{2}+7\times(-1)+7=1 - 7 + 7=1$.

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:

B. 1