Limits
A limit asks where a function is heading, not where it arrives. That distinction is the whole subject: a function can head somewhere it never reaches, and it can arrive somewhere it was not heading. Everything after this — the derivative most of all — is built on limits, so the part worth getting fluent is not the definition but the order of techniques: substitute first, and only do algebra when substituting tells you to.
Calculus — limits and continuity
Graphing calculator- The notationlimx → a f(x) = L
As x gets close to a — from either side, without ever being a — the value of f(x) gets close to L.
- The value at the point is irrelevantlimx → a f(x) need not equal f(a)
The limit looks at the approach, never the arrival. x2−9x−3 has no value at all at x = 3, and its limit there is still 6.
- One-sided limitslimx → a− f(x) and limx → a+ f(x)
Approaching from below and from above. The minus and plus are about which side of a you come from, never about sign.
- When the two-sided limit existslimx → a f(x) = L ⇔ limx → a− f(x) = limx → a+ f(x) = L
Both sides must exist and agree. If they disagree the limit does not exist, however well behaved each side is on its own.
- Continuityf is continuous at a ⇔ limx → a f(x) = f(a)
Heading there and arriving there. Polynomials, exponentials, sines and cosines are continuous everywhere, which is what makes substitution the first thing to try.
- Limits split over the arithmeticlim (f ± g) = lim f ± lim g, lim (fg) = lim f · lim g
And the same for a quotient, provided the bottom limit is not zero. That proviso is where every interesting problem in this lesson lives.
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1. Substitute. Put a in. If you get an ordinary number, that is the answer and you are done — most limits are this easy. Only if substituting breaks do you need anything else, and what it gives you says which.
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2. Got 00 ? Do algebra. This is the indeterminate form: it means "not enough information yet", not "no limit". Factor and cancel, multiply by a conjugate, or combine fractions — whichever the expression invites — then substitute again.
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3. Got k0 with k ≠ 0 ? It is infinite. Not indeterminate at all, and no algebra will fix it. The size is settled — it blows up — so the only question left is the sign, which you get from each side separately.
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4. Two pieces each blowing up? Put them over one denominator. A difference of fractions that each run away is the ∞ − ∞ form, and it is indeterminate: the two infinities can cancel to anything at all. You cannot take the two limits separately. Combine into a single fraction first, then read what that one fraction gives — usually k0, occasionally 00.
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5. Heading to ∞ ? Compare the degrees. Divide top and bottom by the highest power of x in the bottom, or read the degrees off directly with the shortcut below.
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6. A piecewise rule or an absolute value? Take the sides separately. Either one changes its formula at a point, so the two sides are genuinely different functions there. Work out each one-sided limit and then compare them.
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7. Still stuck? Look for a standard limit or a squeeze. A sin shape usually wants the standard trig limit; something bounded multiplied by something heading to zero usually wants the squeeze theorem.
- Zero over zero is indeterminate00 ⇒ do more algebra
It carries no information on its own. The answer could be anything, and finding it is your job.
- Non-zero over zero is infinitek0 ⇒ ±∞, k ≠ 0
Not indeterminate. A fixed non-zero amount divided by something vanishing grows without bound, and no algebra changes that. Only the sign is still open.
- So check the numerator firstbottom → 0 ⇒ what does the top do?
Top also zero means factor and cancel. Top non-zero means infinite. Skipping this check is how x2−4x−2 gets called infinite when its limit is 4.
- Infinite means the limit does not existlimx → a f(x) = ∞ ⇒ no limit, but we say how it fails
∞ is not a number the function approaches. Writing it is a description of the failure — more useful than "does not exist", and still a failure.
- Getting the signtest a point just below a , then just above
Track the sign of the top and the sign of the bottom separately. Near x = 3, x − 3 is a small negative below and a small positive above, which is what flips the answer.
- Opposite sides, no limit at alllimx → 0−1x = −∞, limx → 0+1x = +∞
Disagreeing sides mean the two-sided limit does not exist and cannot be written as ∞ either. An even power underneath, like 1x2, makes both sides agree.
- This is what a vertical asymptote isx = a is a vertical asymptote ⇔ one side is ±∞
The asymptote from the Graphs lesson, said in the language of limits. One side being infinite is enough.
- The methoddivide top and bottom by the highest power of x in the bottom
Every term of the form cxn then heads to zero, and what is left is the answer. This always works; the shortcut below is just it done in advance.
- Bottom winsdeg(top) < deg(bottom) ⇒ 0
Horizontal asymptote y = 0. The bottom outgrows the top, so the fraction is squeezed to nothing.
- A tieaxn + …bxn + … ⇒ ab
Horizontal asymptote at that ratio. Only the leading coefficients matter — everything below the top degree is dust at this scale.
- Top winsdeg(top) > deg(bottom) ⇒ ±∞
No horizontal asymptote. The sign comes from the leading coefficients and the direction x is heading.
- The radical trap√x2 = |x|, which is −x when x < 0
Pulling x out of a root heading to −∞ costs a minus sign. This is the single most missed step in the topic, and it is the Radicals lesson's even-index rule showing up again.
- Standard trig limitlimx → 0 sin xx = 1
In radians only. Also limx → 0 1 − cos xx = 0. These two are what make the derivatives of sine and cosine come out the way they do.
- Squeeze theoremg ≤ f ≤ h and lim g = lim h = L ⇒ lim f = L
For a bounded thing multiplied by a vanishing thing. x2sin1x is trapped between −x2 and x2, so it is dragged to 0.
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The sides disagree — a jump. Each one-sided limit exists and they are different numbers. Typical of a piecewise rule, or of |x|x at zero, which is −1 on the left and 1 on the right.
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It grows without bound — infinite. Written ±∞ when both sides agree. The limit still does not exist; the notation just says how.
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It oscillates. sin1x as x → 0 swings between −1 and 1 infinitely often and settles nowhere. Rare in exercises, but it is why "the function is bounded" is not enough to guarantee a limit.
- 00 ⇒ the limit does not exist00 ⇒ keep going
It is indeterminate, which means undecided rather than impossible. x2−9x−3 gives 00 and has the perfectly ordinary limit 6.
- 20 ⇒ indeterminate20 ⇒ ±∞
Only zero over zero is indeterminate. A fixed non-zero top over a vanishing bottom has a settled size — it blows up — and only the sign is left to find.
- limx → 01x = ∞Does not exist: −∞ on the left, +∞ on the right
Writing a single infinity claims both sides do the same thing. Here they do the opposite. Compare 1x2, where the square makes both sides positive and the single statement is fair.
- limx → −∞ √x2+1x = 1limx → −∞ √x2+1x = −1
√x2 is |x|, not x. Heading to −∞ makes that −x, and the minus survives into the answer.
- limx → 3 f(x) = f(3)Only when f is continuous at 3
That equality is the definition of continuity, not a general fact. Substitution works because most functions you meet are continuous, and the whole interesting part of this lesson is the cases where it does not.
- limx → 0 sin 5xx = 1limx → 0 sin 5xx = 5
The standard limit needs the bottom to match what is inside the sine. Here it does not, and the factor of 5 needed to make it match is exactly what is left over.
Type an answer and it gets marked. What you get right is kept on this device.
- 1.limx → 4 x2 − 16x − 4
- 2.limx → 0 √x + 9 − 3x
- 3.limx → 5 x − 5x2 − 25
- 4.limx → 2+ 3x − 2
- 5.limx → −∞ 4x3 − x2x3 + 7
- 6.limx → ∞ 5x + 2x2 + 1
- 7.limx → 0 sin 7xx
- 8.limx → −∞ √9x2 + 2x
- 9.limx → 1 1x − 1 − 2x2 − 1
- 10.limx → 2− 3x − 2 − 1x2 − 4