If x^(4) 1, at what points c do you automatically know lim_(x rarr c)f(x) ? What can you say about the value of the limit at these points?

Given Inequality: We are given that x^(4) <= f(x) <= x^(2) for x in [-1,1]. This means that as x approaches any c within [-1,1], f(x) is squeezed between x^(4) and x^(2), which both approach c^(4) and c^(2) respectively.Squeeze Theorem Application: Since both x^(4) and x^(2) are continuous functions and they are equal at x = c for any c in [-1,1], by the Squeeze Theorem, the limit of f(x) as x approaches c is also c^(2).Limit Determination: For x 1, we have x^(2) <= f(x) <= x^(4). However, this does not help us determine the limit at any specific point c because the functions x^(2) and x^(4) do not approach the same value as x approaches any c outside of [-1,1].Conclusion: Therefore, we automatically know the limit of f(x) as x approaches any c within the interval [-1,1], and the value of the limit is c^(2).

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If
x^(4) <= f(x) <= x^(2) for
x in
[-1,1] and
x^(2) <= f(x) <= x^(4) for
x < -1 and
x > 1, at what points
c do you automatically know
lim_(x rarr c)f(x) ? What can you say about the value of the limit at these points?

If $x^{4} \leq f(x) \leq x^{2}$ for $x$ in $[-1,1]$ and $x^{2} \leq f(x) \leq x^{4}$ for x < -1 and x > 1, at what points $c$ do you automatically know $\lim_{x \rightarrow c}f(x)$ ? What can you say about the value of the limit at these points?

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Q. If

x^{4} \leq f(x) \leq x^{2}

for

x

[-1,1]

and

x^{2} \leq f(x) \leq x^{4}

for

x < -1

and

x > 1

, at what points

c

do you automatically know

\lim_{x \rightarrow c}f(x)

? What can you say about the value of the limit at these points?

Given Inequality: We are given that $x^{4} \leq f(x) \leq x^{2}$ for $x$ in $[-1,1]$ . This means that as $x$ approaches any $c$ within $[-1,1]$ , $f(x)$ is squeezed between $x^{4}$ and $x^{2}$ , which both approach $c^{4}$ and $x$ $0$ respectively.
Squeeze Theorem Application: Since both $x^{4}$ and $x^{2}$ are continuous functions and they are equal at $x = c$ for any $c$ in $[-1,1]$ , by the Squeeze Theorem, the limit of $f(x)$ as $x$ approaches $c$ is also $c^{2}$ .
Limit Determination: For x < -1 and x > 1, we have $x^{2} \leq f(x) \leq x^{4}$ . However, this does not help us determine the limit at any specific point $c$ because the functions $x^{2}$ and $x^{4}$ do not approach the same value as $x$ approaches any $c$ outside of $[-1,1]$ .
Conclusion: Therefore, we automatically know the limit of $f(x)$ as $x$ approaches any $c$ within the interval $[-1,1]$ , and the value of the limit is $c^{2}$ .