Limits and Continuity

It exists but cannot be determined without further information about f near c from above.

It equals M because M is greater than or equal to f(x).

It equals m since m is less than or equal to f(x).

It does not exist because m is less than or equal to f(x).

1

0

1/2

Undefined

For all x near a, if we have $h(x) \leq f(x) \leq g(x)$ then we need $\lim_{x \to a} h(x) = \lim_{x \to a} g(x) = L$

For any values of x, if we have $(f[object Object]h)(a)$ then we need $\lim_{x \rightarrow a} (f[object Object]h)(a)$

For all x near a, if we have $f(a) \leq g(a) \leq h(a)$ then we need $\lim_{x \to a} h(a) = L$

When approaching any limit L, having inequalities like $(f-g-h)(a) < (g-h-f)(a) < (h-f-g)(a)$ suffices for squeeze application

The function $h(x)$ is undefined at $x=0$, so the squeeze theorem does not apply.

The limits of $f(x)$ and $g(x)$ have no relation to the limit of $h(x)$.

$h(x)$ can be squeezed between $f(x)$ and $g(x)$, showing $\lim_{x \to 0} h(x) = 0$.

Since $\sin\left( \frac{1}{x} \right)$ oscillates, the squeeze theorem cannot apply to $h(x)$.

If two functions are continuous, their limits are always zero.

If two functions have the same limit, their derivatives are also equal.

If two functions are bounded by a third function, their limits are also bounded.

If two functions have the same derivative, their limits are also equal.

0

Infinity

Undefined

1

Only integer values for k make the statement true due to periodic properties of cosine.

Only multiples of $\pi$ for k make this true to keep cosine from oscillating too much.

Any real number value for k makes the statement true because cosine values are bounded between -1 and 1.

There is no such k because cosine does not have limits at infinity.

It must also equal to zero.

It must equal to L.

It depends on the value of k.

It does not exist.

$\lim_{x \to -\infty} h(x)=+\infty$, Because x to the fourth power approaches positive infinity as x approaches negative infinity

$\lim_{x \to -\infty} h(x)=0$, since both bounding functions approach zero

$\lim_{x \to -\infty} h(x)=-\infty$, Since x to the fourth power approaches negative infinity as x approaches negative infinity

No conclusion can be drawn because while sine squared approaches zero, x to the fourth power does not have an upper bound as x approaches negative infinity.

$\lim_{x \to 0} \sin(x) = 0$

$\lim_{x \to \infty} \sin(x) = 0$

There is not enough information to determine $\lim_{x \to 0} \sin(x)$

$\lim_{x \to 0} \sin(x) = -1$