Question

let (f), (g), and (h) be the functions defined by (f(x)=\frac{1 - cos x}{x^{2}}), (g(x)=x^{2}sin(\frac{1}{x})), and (h(x)=\frac{sin x}{x}) for (x
eq0). all of the following inequalities are true 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}{2}(1 - x^{2})leq f(x)leq\frac{1}{2})
ii. (-x^{2}leq g(x)leq x^{2})
iii. (-\frac{1}{|x|}leq h(x)leq\frac{1}{|x|})

Explanation:

Step1: Recall squeeze - theorem conditions

The squeeze theorem states that if \(l(x)\leq y(x)\leq u(x)\) for all \(x\) in some open interval containing \(a\) (except possibly at \(x = a\)) and \(\lim_{x
ightarrow a}l(x)=\lim_{x
ightarrow a}u(x)=L\), then \(\lim_{x
ightarrow a}y(x)=L\). We need to check the limits of the lower - and upper - bound functions as \(x
ightarrow0\).

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

For \(f(x)\), consider \(\lim_{x
ightarrow0}\frac{1}{3}(1 - x^{2})\) and \(\lim_{x
ightarrow0}\frac{1}{2}\).
\(\lim_{x
ightarrow0}\frac{1}{3}(1 - x^{2})=\frac{1}{3}(1-0)=\frac{1}{3}\) and \(\lim_{x
ightarrow0}\frac{1}{2}=\frac{1}{2}\). Since \(\lim_{x
ightarrow0}\frac{1}{3}(1 - x^{2})
eq\lim_{x
ightarrow0}\frac{1}{2}\), we cannot use the squeeze theorem with this inequality for \(f(x)\) as \(x
ightarrow0\).

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

For \(g(x)=x^{2}\sin(\frac{1}{x})\), consider \(\lim_{x
ightarrow0}-x^{2}\) and \(\lim_{x
ightarrow0}x^{2}\).
We know that \(\lim_{x
ightarrow0}-x^{2}=0\) and \(\lim_{x
ightarrow0}x^{2}=0\). Since \(-x^{2}\leq x^{2}\sin(\frac{1}{x})\leq x^{2}\) and \(\lim_{x
ightarrow0}-x^{2}=\lim_{x
ightarrow0}x^{2}=0\), by the squeeze theorem, \(\lim_{x
ightarrow0}x^{2}\sin(\frac{1}{x}) = 0\).

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

For \(h(x)=\frac{\sin x}{x}\), consider \(\lim_{x
ightarrow0}-\frac{1}{|x|}\) and \(\lim_{x
ightarrow0}\frac{1}{|x|}\).
\(\lim_{x
ightarrow0}-\frac{1}{|x|}=-\infty\) and \(\lim_{x
ightarrow0}\frac{1}{|x|}=\infty\). Since the limits of the lower - and upper - bound functions are not equal as \(x
ightarrow0\), we cannot use the squeeze theorem with this inequality for \(h(x)\) as \(x
ightarrow0\).

Answer:

II only