Supplemental Math Project · Calculus
A short tutorial — solving two limits, each by two independent methods, and showing the answers agree.
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The idea
A limit asks a simple question: as x gets closer and closer to some number, what value is the function heading toward?
"As x approaches a, f(x) approaches L."
x never has to equal a — we only care where the function is going. That is why a limit can exist even when plugging in the number gives 0/0.
Problem 1 of 2
We will evaluate it two ways — first by algebra (factoring & simplifying), then by a numerical table — and check that both give the same answer.
Problem 1 · why we cannot just plug in
Plugging in x = 2 gives 0 over 0.
0/0 is not the answer and not "undefined for good" — it is a signal that we must simplify first, then substitute.
Problem 1 · Method 1
The numerator is a difference of squares: a² − b² = (a − b)(a + b).
Rewrite the fraction with the factored top.
Cancel the common factor (x − 2). Allowed because near the limit x ≠ 2, so x − 2 ≠ 0.
Now substitution is safe — the simplified function is continuous.
Problem 1 · Method 2
A limit is about where the output heads. So feed in x-values creeping toward 2 from both sides and watch the output.
| x → | 1.9 | 1.99 | 1.999 | 2.001 | 2.01 | 2.1 |
|---|---|---|---|---|---|---|
| f(x) | 3.9 | 3.99 | 3.999 | 4.001 | 4.01 | 4.1 |
| — | approaching from the left → 4 | 4 ← approaching from the right | ||||
From the left and the right the outputs close in on 4. The table confirms the algebra.
Problem 1 · Result
Factoring gave 4. The table closed in on 4. Two independent methods, one answer — so we can trust it.
Problem 2 of 2
A fraction stacked inside a fraction — a "complex fraction."
Same plan: simplify the messy algebra first (Method 1), then verify with a table (Method 2).
Problem 2 · Method 1
Direct substitution is indeterminate again — simplify first.
Combine the top two fractions over the common denominator 2(x + 2).
Dividing by x is multiplying by 1/x. The x cancels — that was the troublemaker.
Now substitute x = 0 safely.
Problem 2 · Method 2
Evaluate the original expression at x-values squeezing toward 0 from both sides.
| x → | −0.1 | −0.01 | −0.001 | 0.001 | 0.01 | 0.1 |
|---|---|---|---|---|---|---|
| f(x) | −0.2632 | −0.2513 | −0.2501 | −0.2499 | −0.2488 | −0.2381 |
| — | heading toward −0.25 | −0.25 ← heading toward | ||||
Both sides converge on −0.25, which is exactly −1/4.
Problem 2 · Result
Simplifying gave −1/4. The table closed in on −0.25. Same value — answer confirmed.
Wrapping up
An indeterminate form is a prompt to do algebra — factor or combine — before substituting.
Difference of squares, cancelling a common factor, and combining a complex fraction.
Watching outputs from both sides confirms where the function is heading.
Problem 1 → 4. Problem 2 → −1/4. Both methods agreed in each case.
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