Question

webwork 1 - topics 2 - 3: problem 10 (1 point)
suppose that (a) is a constant and (f(x)) is a function of (x) such that
(\frac{ax}{x - 3}<f(x)<x - 243)
for all (x) near 27 but not equal to 27.
we are interested in finding the limit of (f(x)) as (x) approaches 27 by means of the squeeze theorem.
(a) for the squeeze theorem to be applicable in this case, the constant (a) must be equal to a specific number. find this number.
answer: (a=)
(b) assuming that (a) is that number for which the squeeze theorem is applicable, find (lim_{x
ightarrow27}f(x))
answer: this limit is equal to

Explanation:

Step1: Apply Squeeze - Theorem condition

For the Squeeze Theorem to be applicable, $\lim_{x
ightarrow27}\frac{Ax}{x - 3}=\lim_{x
ightarrow27}(x - 243)$. First, find $\lim_{x
ightarrow27}(x - 243)$.
$\lim_{x
ightarrow27}(x - 243)=27-243=- 216$.

Step2: Find the limit of $\frac{Ax}{x - 3}$ as $x

ightarrow27$
$\lim_{x
ightarrow27}\frac{Ax}{x - 3}=\frac{A\times27}{27 - 3}=\frac{27A}{24}=\frac{9A}{8}$.

Step3: Solve for $A$

Set $\frac{9A}{8}=-216$. Cross - multiply: $9A=-216\times8$. Then $9A=-1728$. Divide both sides by 9: $A = - 192$.

Step4: Find $\lim_{x

ightarrow27}f(x)$
Since $\lim_{x
ightarrow27}\frac{-192x}{x - 3}=\lim_{x
ightarrow27}(x - 243)=-216$ and $\frac{-192x}{x - 3}<f(x)<x - 243$ for $x$ near 27, by the Squeeze Theorem, $\lim_{x
ightarrow27}f(x)=-216$.

Answer:

(a) $A=-192$
(b) $\lim_{x
ightarrow27}f(x)=-216$