Questions: Suppose -x^2+4x-1 ≤ f(x) ≤ x^2-4x+7 for all x values near x=2, except possibly at 2. Evaluate the limit as x approaches 2 of f(x). a.) The limit as x approaches 2 of f(x)=3 b.) The limit as x approaches 2 of f(x) does not exist c.) The limit as x approaches 2 of f(x)=7 d.) The limit as x approaches 2 of f(x)=11

Suppose -x^2+4x-1 ≤ f(x) ≤ x^2-4x+7 for all x values near x=2, except possibly at 2.

Evaluate the limit as x approaches 2 of f(x).
a.) The limit as x approaches 2 of f(x)=3
b.) The limit as x approaches 2 of f(x) does not exist
c.) The limit as x approaches 2 of f(x)=7
d.) The limit as x approaches 2 of f(x)=11

Transcript text: Suppose $-x^{2}+4 x-1 \leq f(x) \leq x^{2}-4 x+7$ for all $x$ values near $x=2$, except possibly at 2 . Evaluate $\lim _{x \rightarrow 2} f(x)$. a.) $\lim _{x \rightarrow 2} f(x)=3$ b.) $\lim _{x \rightarrow 2} f(x)$ does not exist c.) $\lim _{x \rightarrow 2} f(x)=7$ d.) $\lim _{x \rightarrow 2} f(x)=11$

Solution

Evaluate $\lim _{x \rightarrow 2} f(x)$ using the given inequality $-x^{2}+4x-1 \leq f(x) \leq x^{2}-4x+7$.

Find the limit of the lower bound function as $x \rightarrow 2$.

Evaluate $\lim _{x \rightarrow 2} (-x^{2}+4x-1)$:
\[ \lim _{x \rightarrow 2} (-x^{2}+4x-1) = -(2)^{2} + 4(2) - 1 = -4 + 8 - 1 = 3.
\]

Find the limit of the upper bound function as $x \rightarrow 2$.

Evaluate $\lim _{x \rightarrow 2} (x^{2}-4x+7)$:
\[ \lim _{x \rightarrow 2} (x^{2}-4x+7) = (2)^{2} - 4(2) + 7 = 4 - 8 + 7 = 3.
\]

Apply the Squeeze Theorem.

Since $-x^{2}+4x-1 \leq f(x) \leq x^{2}-4x+7$ and both bounds approach 3 as $x \rightarrow 2$, by the Squeeze Theorem, $\lim _{x \rightarrow 2} f(x) = 3$.

$\boxed{\lim _{x \rightarrow 2} f(x) = 3}$

$\boxed{\lim _{x \rightarrow 2} f(x) = 3}$
The answer is A.

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