webwork 1 - topics 2 - 3: problem 10 (1 point)\nsuppose that (a) is a constant and (f(x)) is a function of…

webwork 1 - topics 2 - 3: problem 10 (1 point)\nsuppose that (a) is a constant and (f(x)) is a function of (x) such that\n(\frac{ax}{x - 3}<f(x)<x - 243)\nfor all (x) near 27 but not equal to 27.\nwe are interested in finding the limit of (f(x)) as (x) approaches 27 by means of the squeeze theorem.\n(a) for the squeeze theorem to be applicable in this case, the constant (a) must be equal to a specific number. find this number.\nanswer: (a=)\n(b) assuming that (a) is that number for which the squeeze theorem is applicable, find (lim_{x\rightarrow27}f(x))\nanswer: this limit is equal to
Answer
Explanation:
Step1: Apply Squeeze - Theorem condition
For the Squeeze Theorem to be applicable, $\lim_{x\rightarrow27}\frac{Ax}{x - 3}=\lim_{x\rightarrow27}(x - 243)$. First, find $\lim_{x\rightarrow27}(x - 243)$. $\lim_{x\rightarrow27}(x - 243)=27-243=- 216$.
Step2: Find the limit of $\frac{Ax}{x - 3}$ as $x\rightarrow27$
$\lim_{x\rightarrow27}\frac{Ax}{x - 3}=\frac{A\times27}{27 - 3}=\frac{27A}{24}=\frac{9A}{8}$.
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 $\lim_{x\rightarrow27}f(x)$
Since $\lim_{x\rightarrow27}\frac{-192x}{x - 3}=\lim_{x\rightarrow27}(x - 243)=-216$ and $\frac{-192x}{x - 3}<f(x)<x - 243$ for $x$ near 27, by the Squeeze Theorem, $\lim_{x\rightarrow27}f(x)=-216$.
Answer:
(a) $A=-192$ (b) $\lim_{x\rightarrow27}f(x)=-216$