Question

suppose -5x + 6 ≤ f(x) ≤ x² - 7x + 7. use this to compute the following limit. lim(x→1) f(x). answer: blank. what theorem did you use to arrive at your answer? answer: blank.

Explanation:

Step1: Find left - hand limit

Compute $\lim_{x
ightarrow1}(-5x + 6)$. Substitute $x = 1$ into $-5x+6$:
$\lim_{x
ightarrow1}(-5x + 6)=-5\times1 + 6=1$

Step2: Find right - hand limit

Compute $\lim_{x
ightarrow1}(x^{2}-7x + 7)$. Substitute $x = 1$ into $x^{2}-7x + 7$:
$\lim_{x
ightarrow1}(x^{2}-7x + 7)=1^{2}-7\times1 + 7=1$

Step3: Apply Squeeze Theorem

Since $-5x + 6\leq f(x)\leq x^{2}-7x + 7$ and $\lim_{x
ightarrow1}(-5x + 6)=\lim_{x
ightarrow1}(x^{2}-7x + 7)=1$, by the Squeeze Theorem, $\lim_{x
ightarrow1}f(x)=1$

Answer:

1
Squeeze Theorem