Question

let $f$, $g$, and $h$ be the functions defined by $f(x)=\frac{sin x}{2x}$, $g(x)=x^{4}cos(\frac{1}{x^{2}})$, and $h(x)=\frac{x^{2}}{\tan x}$ for $x
eq0$. all of the following inequalities are true on the interval $-1,1$ for $x
eq0$. which of the inequalities can be used with the squeeze theorem to find the limit of the function as $x$ approaches $0$?
i. $\frac{1}{4}leq f(x)leq x^{2}+\frac{1}{2}$
ii. $-x^{4}leq g(x)leq x^{4}$
iii. $-\frac{1}{x^{2}}leq h(x)leq\frac{1}{x^{2}}$

Explanation:

Step1: Recall squeeze - theorem conditions

The squeeze theorem states that if \(l(x)\leq y(x)\leq u(x)\) for all \(x\) in an open - interval containing \(a\) (except possibly at \(x = a\)) and \(\lim_{x
ightarrow a}l(x)=\lim_{x
ightarrow a}u(x)=L\), then \(\lim_{x
ightarrow a}y(x)=L\).

Step2: Analyze inequality I for \(f(x)\)

\(\lim_{x
ightarrow0}\frac{1}{4}=\frac{1}{4}\) and \(\lim_{x
ightarrow0}(x^{2}+\frac{1}{2})=\frac{1}{2}\). Since \(\lim_{x
ightarrow0}\frac{1}{4}
eq\lim_{x
ightarrow0}(x^{2}+\frac{1}{2})\), we cannot use the squeeze theorem with the inequality \(\frac{1}{4}\leq f(x)\leq x^{2}+\frac{1}{2}\).

Step3: Analyze inequality II for \(g(x)\)

We know that \(- 1\leq\cos(\frac{1}{x^{2}})\leq1\) for \(x
eq0\). Then \(-x^{4}\leq x^{4}\cos(\frac{1}{x^{2}})\leq x^{4}\). Also, \(\lim_{x
ightarrow0}(-x^{4}) = 0\) and \(\lim_{x
ightarrow0}(x^{4}) = 0\). By the squeeze theorem, \(\lim_{x
ightarrow0}g(x)=\lim_{x
ightarrow0}x^{4}\cos(\frac{1}{x^{2}})=0\).

Step4: Analyze inequality III for \(h(x)\)

\(\lim_{x
ightarrow0}(-\frac{1}{x^{2}})=-\infty\) and \(\lim_{x
ightarrow0}(\frac{1}{x^{2}})=\infty\). Since the limits of the lower - and upper - bound functions are not equal and finite as \(x
ightarrow0\), we cannot use the squeeze theorem with the inequality \(-\frac{1}{x^{2}}\leq h(x)\leq\frac{1}{x^{2}}\).

Answer:

II only