Unit 1: Limits and Continuity

• Finding speed at an instantaneous velocity

• x approaches but never reaches a from : -the left limx→a−​f(x) - the right limx→a+​f(x)

• The left and right limits must be the same for the limit to exist

How to find the limit?

1. Direct substitution. Plug in the number that the limit is approaching.

An example of direct substitution in finding a limit can be illustrated with the function $f(x) = 2x + 3$. To find the limit as $x$ approaches 4, we perform direct substitution:

$\lim_{x \to 4} (2x + 3)$

By substituting 4 into the function:

$= 2(4) + 3$
$= 8 + 3$
$= 11$

Thus, $\lim_{x \to 4} (2x + 3) = 11$.

2. If you get a denominator of 0, try factoring the numerator and denominator, cancelling out the common factor! (refer to notes for example)

3. If you cannot factor, you have to look at the limits from both the left and the right side; then the limit does not exist. This is because the denominator is getting infinitely smaller as a gets approached from the left and it gets infinitely bigger as a gets approached from the right. (refer to notes for example)

4. HOWEVER: If you cannot factor, but the denominator is squared, the limit will approach positive infinity from both the left and the right because anything squared is positive! (refer to notes for example)

5. When x approaches infinity, you can cancel out everything but the coefficient attached to it. (refer to notes for example)

6. If the limit approaches 0, you have to solve the heck out of it algebraically

Squeeze theorem: The Squeeze Theorem (Sandwich Theorem)

Definition:given $g\left(x\right)\le f\left(x\right)\le h\left(x\right)$ , if g(x) and h(x) are equal and have the same limit, f(x) will also be equal.

• $\lim_{x\to a}$ (constant)= constant

• Given a piecewise function, both limits have to match or else the limit will not exist.

Trig functions to memorize (LIMIT MUST APPROACH 0)

$\lim_{x\to0}\frac{\left(\sin x\right)}{x}=1$

$\lim_{x\to0}\frac{\left(1-\cos x\right)}{x}=0$

$\lim_{x\to0}\frac{x}{\sin x}=1$

$\lim_{x\to0}\frac{\left(\cos x-1\right)}{x}=0$

Continuity

• If you can trace a graph without lifting a pencil, the function is continuous

Discontinuity:

• Removable discontinuity (holes): where numerator and denominator cancel

• Vertical Asymptote: the denominator that does not cancel out

• Jump: When the piecewise function limit values do not match.

Continuity at one point

1. $f\left(c\right)$ is defined

2. $\lim_{x\to c}f\left(x\right)$ exists

3. $\lim_{x\to c}f\left(x\right)=f\left(c\right)$

Intermediate Value Theorem

• k is a value between [a,b]

1. Confirm function is continuous on [a,b]

2. Make sure f(a) does not equal f(b)

3. check that k is between f(a) and f(b)

• f(x)=k will exist between the interval [a,b]

Miscellaneous

• average rate of change is slope

• Instaneous rate of change is slope of tangent line

• steep means going at a faster rate and bigger slope ( more vertical than horizontal