The Big Idea: Instantaneous Rate of Change#

In everyday life, we talk about average speed (like “I drove 120 miles in 2 hours, so my average speed was 60 mph”). But what about your speed at a single instant (the number your car’s speedometer shows at exactly t = 30 seconds)?

In algebra, the slope of a line $\Delta y / \Delta x$ measures average rate of change between two points. In calculus, we want the slope of a curve (the instantaneous rate of change) at exactly one point.

To handle this, we let the two points on the curve get very close—so close that the difference between their $x$-coordinates ($\Delta x$) goes to zero. Limits formalize the idea of “approaching a value” without necessarily reaching it.

Left-Hand vs. Right-Hand Limits#

• Left-hand limit: $\lim_{x \to a^-} f(x)$ means $x$ approaches $a$ from values less than $a$.
• Right-hand limit: $\lim_{x \to a^+} f(x)$ means $x$ approaches $a$ from values greater than $a$.

For $\lim_{x \to a} f(x)$ to exist (and be equal to some $L$), the left-hand limit and right-hand limit must agree.

Estimating from a Graph#

1. Look at the point $x=a$ on your graph.
2. Trace from the left side and the right side to see if the function values head toward the same $y$-value.
3. If they do, that $y$-value is the limit.
4. If they approach two different $y$-values, the limit does not exist (DNE).

Note: The function might have a hole or no actual point at $x=a$. The limit can still exist if the $y$-value approached from both sides is the same.

Estimating from a Table#

If you have a table of $(x, f(x))$ values near $x = a$, look at how $f(x)$ behaves as $x$ gets closer to $a$ from above and below. If the outputs get closer to one number, that number is your estimated limit.

Squeeze (Sandwich) Theorem#

If $g(x)\leq f(x)\leq h(x)$ for $x$ near $a$, and if

then

Example: $\lim_{x \to 0} x \sin\left(\frac{1}{x}\right)$.

• We know $-1 \le \sin\left(\frac{1}{x}\right) \le 1$.
• Multiply by $x$: $-|x|\le x\,\sin\left(\tfrac{1}{x}\right)\le |x|$.
• $\lim_{x \to 0} -|x| = 0$ and $\lim_{x \to 0} |x| = 0$.
• By the Squeeze Theorem, the limit is $0$.

Continuity at a Point#

A function $f$ is continuous at $x=a$ if:

1. $f(a)$ is defined.
2. $\lim_{x \to a} f(x)$ exists.
3. $\lim_{x \to a} f(x) = f(a)$.

In simpler terms, you can draw the function at $x=a$ without lifting your pencil.

Types of Discontinuities#

1. Removable (Hole): The limit exists, but $f(a)$ is not defined or isn’t equal to that limit.
2. Jump: Left- and right-hand limits exist but are not equal.
3. Infinite: The function goes to $+\infty$ or $-\infty$ near $a$, usually indicating a vertical asymptote at $x=a$.

Removing a Discontinuity#

Sometimes you can redefine the function at $a$ to “fill in” a hole, making it continuous.

Infinite Limits (Vertical Asymptotes)#

means $f(x)$ becomes arbitrarily large (positive) as $x$ approaches $a$. Often, $x=a$ is a vertical asymptote. Similarly for $-\infty$.

Limits at Infinity (Horizontal Asymptotes)#

To understand how a function behaves as $x$ grows large (positively or negatively), look at:

If a function approaches a finite value $L$ as $x\to\infty$, then $y=L$ is a horizontal asymptote.

Common Trick: For rational functions $\frac{P(x)}{Q(x)}$:

• If $\deg(P)=\deg(Q)$, limit at infinity is ratio of leading coefficients.
• If $\deg(P)<\deg(Q)$, the limit is $0$.
• If $\deg(P)>\deg(Q)$, the function grows without bound and no horizontal asymptote exists (there might be a slant asymptote instead).

Self-Check 1#

Problem: Evaluate the limit by direct substitution if possible, or determine if it does not exist:

Solution 1#

1. $2x+1$ is a polynomial, continuous everywhere.
2. Substitute $x=3$: $$2(3)+1 = 7.$$

So, $\boxed{7}$ is the limit.

Self-Check 2#

Problem: Evaluate:

Hint: You get $\frac{0}{0}$ if you plug in $x=1$ directly, so consider factoring.

Solution 2#

1. Factor numerator: $x^2 - 1 = (x-1)(x+1)$.
2. Rewrite the expression: $$\frac{(x-1)(x+1)}{x-1}.$$ For $x \neq 1$, the $(x-1)$ cancels.
3. So the limit depends on $\lim_{x \to 1} (x+1)$, which is $2$.
4. $\boxed{2}$ is the limit.

Self-Check 3#

Problem: Use the Squeeze Theorem to find:

Solution 3#

1. $-1 \leq \sin\bigl(\tfrac{1}{x}\bigr)\leq 1$.
2. Multiply by $x$: $-|x|\le x\,\sin\bigl(\tfrac{1}{x}\bigr)\le |x|$.
3. As $x\to 0$, both $-|x|$ and $|x|$ go to 0.
4. By Squeeze Theorem, the limit is $\boxed{0}$.

Self-Check 4#

Problem: Classify the discontinuity at $x=2$ for

Then find $\lim_{x \to 2} f(x)$.

Solution 4#

1. For $x \neq 2$: $$\frac{x^2 - 4}{x - 2} = \frac{(x-2)(x+2)}{x-2} = x+2 \quad (x\neq 2).$$
2. As $x\to 2$, $x+2 \to 4$. So the limit from either side is 4.
3. But $f(2)=10$. The limit (4) is not equal to the function’s defined value (10).
4. This is a removable discontinuity (a “hole”) at $x=2$.
5. $\lim_{x \to 2} f(x)=\boxed{4}$.

Self-Check 5#

Problem: Evaluate:

Solution 5#

1. Factor out $x^2$ from numerator and denominator:

   $$\frac{3x^2 + 5x -2}{6x^2 -4} = \frac{x^2(3 + 5/x -2/x^2)}{x^2(6 - 4/x^2)} = \frac{3 + 5/x -2/x^2}{6 -4/x^2}.$$

2. As $x\to\infty$, $5/x, 2/x^2,$ and $4/x^2$ all go to 0.

3. The expression approaches $\frac{3}{6} = \frac12$.

4. $\boxed{\tfrac12}$ is the limit. A horizontal asymptote is $y=\tfrac12$.

Self-Check 6#

Problem: Show by the Intermediate Value Theorem that

has at least one real root in $[0,1]$.

Solution 6#

1. $g(0)= -1$. $g(1)= 1$.
2. $g$ is a polynomial, so it’s continuous on $[0,1]$.
3. $g(0)<0$ and $g(1)>0$.
4. By IVT, there is some $c$ in $[0,1]$ for which $g(c)=0$. Hence a root exists in that interval.