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What a limit says, why the value at the point is irrelevant, and when the two sides fail to agree.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to say what $\lim_{x \to a} f(x) = L$ claims and what it deliberately says nothing about, compute a limit that needs one algebraic step — factoring, rationalising or combining a compound fraction — compute one-sided limits at a junction or an absolute value and decide from them whether the two-sided limit exists, and distinguish the limit at a point from the value at that point in every combination the two can occur in.
$\lim_{x \to a} f(x) = L$ means: the values of $f$ can be made as close to $L$ as you like by taking $x$ close enough to $a$, without ever taking $x = a$.
That last clause is the whole idea. A limit says what the outputs approach, not what the function does at the point. $f(a)$ may differ from $\lim_{x \to a} f(x)$, and it may not exist at all, without the limit being affected. A limit is a statement about a punctured neighbourhood of $a$ — the interval round $a$ with $a$ itself lifted out — and the function's behaviour at the missing point is not part of the question.
This is not a technicality. The central object of the next unit is $$\lim_{h \to 0} \frac{f(a + h) - f(a)}{h},$$ and at $h = 0$ that quotient is $0/0$: it has no value at the point, ever. A notion of limit that consulted the value would be useless for it.
One-sided limits. $\lim_{x \to a^-}$ approaches from below, $\lim_{x \to a^+}$ from above, and $$\lim_{x \to a} f(x) = L \iff \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L.$$ Both must exist and agree. Existing separately is not enough, and this is where most "the limit does not exist" answers come from.
Another way: picture
Cover the point $a$ with your thumb and look at the graph either side of it. If the two edges of what you can see head for the same height, that height is the limit; if they head for different heights, there is none. Now lift your thumb: whatever is underneath — a dot in the right place, a dot in the wrong place, or no dot at all — changes nothing you just decided.
Another way: steps
Substituting and getting $0/0$ tells you exactly one thing: the limit is not being decided by substitution. It is called an indeterminate form because every possible outcome is still on the table.
| Limit | Value |
|---|---|
| $\lim_{x \to 0} \dfrac{x^2}{x}$ | $0$ |
| $\lim_{x \to 0} \dfrac{3x}{x}$ | $3$ |
| $\lim_{x \to 0} \dfrac{x}{x^2}$ | does not exist (unbounded) |
| $\lim_{x \to 0} \dfrac{\|x\|}{x}$ | does not exist (the sides disagree) |
All four substitute to $0/0$. The form is the same and the answers are different, so the form cannot be the answer. What settles it is how fast the two vanishings happen relative to each other, and the algebra — factoring, rationalising, combining — is how that is made visible.
By contrast, a non-zero number over zero is not indeterminate: $\frac{3}{0^+} \to +\infty$ and $\frac{3}{0^-} \to -\infty$. The size is settled; only the sign needs a one-sided look.
Writing $\lim_{x \to a} f(x) = f(a)$ by habit. That equation is the definition of continuity, not a fact about limits, and the interesting limits are exactly the ones where it fails or is not even well posed.
Treating $0/0$ as $0$, or as $1$. It is neither. It is a signal.
Cancelling and then forgetting the cancellation. $\frac{x^2 - 4}{x-2}$ and $x + 2$ are different functions; they agree off $x = 2$, which is all a limit needs, but the original still has no value at $2$.
Taking existence of both one-sided limits as existence of the limit. $\frac{|x|}{x}$ has both, $-1$ and $1$, and no two-sided limit. Agreement is the condition.
Writing $= \infty$ and calling it a limit. In this course a limit is a real number. $\lim_{x\to 0} 1/x^2 = \infty$ is a statement about how the limit fails, and it is worth saying, but it is not a limit existing.
This works often enough to become a habit, and the habit is what fails on every limit worth computing. Substitution gives the limit only for functions that are continuous at the point — which is the theorem of a later lesson, with hypotheses. At a point where the function is not defined, substitution has nothing to substitute into; at a jump it gives one side's answer and hides the other; at a removable discontinuity it gives an answer the neighbouring values contradict. The reliable order of operations is: try substitution, and when it gives $0/0$ or nothing at all, that is the beginning of the problem, not the end of it.
Factor. $\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3} = \lim_{x \to 3} \dfrac{(x-3)(x+3)}{x-3} = \lim_{x \to 3}(x + 3) = 6$.
The vanishing factor cancels.
Rationalise. $\lim_{x \to 4} \dfrac{\sqrt{x} - 2}{x - 4}$: multiply above and below by $\sqrt{x} + 2$ to get $\dfrac{x - 4}{(x-4)(\sqrt x + 2)} = \dfrac{1}{\sqrt x + 2} \to \dfrac14$.
The conjugate turns the root into the vanishing factor.
Combine. $\lim_{x \to 3} \dfrac{1/x - 1/3}{x - 3}$: the numerator is $\dfrac{3 - x}{3x}$, so the quotient is $\dfrac{-(x-3)}{3x(x-3)} = \dfrac{-1}{3x} \to -\dfrac19$.
A compound fraction hides its own vanishing factor.
$f(x) = \dfrac{\sin x}{x}$ is undefined at $x = 0$ — the formula genuinely gives $0/0$ there — and $\lim_{x \to 0} f(x) = 1$, as the next lessons will prove.
No value, and a perfectly good limit.
Now define $g(x) = 1$ for $x \ne 0$ and $g(0) = 5$. Then $\lim_{x \to 0} g(x) = 1$ while $g(0) = 5$.
A value, a limit, and they disagree.
And $\lfloor x \rfloor$ at $x = 2$ has value $2$, right limit $2$, left limit $1$: a value, one side agreeing with it, and no limit at all.
Every combination occurs, which is why they are separate questions.
Substituting gives $0/0$, so factor: $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$.
The difference of cubes.
Cancelling the $x - 2$ leaves $x^2 + 2x + 4$, which is continuous at $2$.
So the limit is $4 + 4 + 4 = 12$.
Match each local graph behavior to the appropriate limit conclusion.
| $\lim_{x\to2^-}f(x)=7$ | $\lim_{x\to2^+}f(x)=7$ | $\lim_{x\to2}f(x)=7$ | The two-sided limit does not exist | |
|---|---|---|---|---|
| As $x$ approaches 2 from below, $f(x)$ approaches 7 | ||||
| As $x$ approaches 2 from above, $f(x)$ approaches 7 | ||||
| Both directional limits equal 7 | ||||
| Directional limits are 1 and 4 |
Match each limit situation to the evidence needed for its conclusion.
| Substitute directly to get 5 | Factor and cancel before substituting | Compare left and right limits | The limit is 4, independently of the assigned value | |
|---|---|---|---|---|
| $\lim_{x\to2}(x^2+1)$ | ||||
| $\lim_{x\to3}\frac{x^2-9}{x-3}$ | ||||
| A piecewise formula changes at $x=1$ | ||||
| $f(2)=99$ but nearby values approach 4 |
Match each description near $x=2$ to the correct conclusion.
| Limit exists and equals 5, but there is no function value | Limit is 5 although the function value is 9 | The two-sided limit does not exist | The limit and function value agree; this is continuous there | |
|---|---|---|---|---|
| Both sides approach 5; $f(2)$ is undefined | ||||
| Both sides approach 5; $f(2)=9$ | ||||
| Left side approaches 1; right side approaches 4 | ||||
| Both sides approach 3 and $f(2)=3$ |
Find $\displaystyle\lim_{x \to 7} \frac{x^2 - 49}{x - 7}$.
Answer:
Find $\displaystyle\lim_{x \to 3} \dfrac{1/x - 1/3}{x - 3}$, to four decimal places where it is not a whole number.
Answer:
$h(x) = 2x$ for $x \le 1$ and $h(x) = 6x + k$ for $x > 1$. For which $k$ does $\displaystyle\lim_{x \to 1} h(x)$ exist?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For $f(x) = \dfrac{|x - 2|}{x - 2}$ at $x = 2$, does $\displaystyle\lim_{x \to 2} f(x)$ exist?
You can clear a $0/0$ by algebra, compute one-sided limits, and say whether a two-sided limit exists. Say in your own words why a limit must ignore the value at the point, and what would break if it did not. Next: the epsilon-delta definition, which makes "as close as you like" precise.
8. Your turn: $\lim_{x \to 2} \dfrac{x^3 - 8}{x - 2}$, step 3