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).

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.

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

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

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

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

2. The left-hand limit and right-hand limit of f(a) are equal.

3. Function f(a) has a removable discontinuity at point a.

There are two other functions g(x) and h(x) such that $g(x) \leq f(x) \leq h(x)$

1. Function f'(a) exists and is continuous at a

1

0

1/2

Undefined

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.

$\lim_{x \to -\infty} h(x)=+\infty$, 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.

$\frac{1}{x}$

Does not exist

$0$

$1$