Introduction to Limits

Study Sheet

Introduction to Limits

The idea of a limit, evaluating limits, one-sided & infinite limits, limits at infinity, continuity, and a first look at the derivative

The Idea of a Limit; Estimating Numerically and Graphically

Concept
What a limit means

The limit of f(x)f(x) as xx approaches aa is the single value LL that the outputs f(x)f(x) get arbitrarily close to as xx gets close to aa (from both sides), without necessarily equaling aa. We write

lim⁡x→af(x)=L.\lim_{x\to a} f(x) = L.

The value f(a)f(a) itself may be different from LL, or may not even exist---the limit only cares about the behavior near aa.

Example
Estimating a limit from a table

Estimate lim⁡x→2x2−4x−2\displaystyle\lim_{x\to 2}\frac{x^2-4}{x-2}. The function is undefined at x=2x=2 (it gives 00\tfrac{0}{0}), so we build a table of nearby inputs:

From both sides the outputs approach 44, so lim⁡x→2x2−4x−2=4\displaystyle\lim_{x\to 2}\frac{x^2-4}{x-2}=4, even though f(2)f(2) is undefined.

Tip

Tip: A hole in a graph does not stop a limit from existing. Read a limit by tracing the curve toward x=ax=a with your finger---the height you approach is LL, whether or not the point is filled in.

Evaluating Limits Algebraically

Concept
The limit laws

If lim⁡x→af(x)\lim\limits_{x\to a}f(x) and lim⁡x→ag(x)\lim\limits_{x\to a}g(x) both exist, then limits distribute over the arithmetic:

lim⁡x→a[f±g]=lim⁡f±lim⁡g,lim⁡x→a[f⋅g]=lim⁡f⋅lim⁡g,\lim_{x\to a}\big[f\pm g\big]=\lim f\pm\lim g,\qquad \lim_{x\to a}\big[f\cdot g\big]=\lim f\cdot\lim g,
lim⁡x→afg=lim⁡flim⁡g  (lim⁡g≠0),lim⁡x→a[f(x)]n=[lim⁡x→af(x)]n.\lim_{x\to a}\frac{f}{g}=\frac{\lim f}{\lim g}\ \ (\lim g\ne 0),\qquad \lim_{x\to a}\big[f(x)\big]^n=\Big[\lim_{x\to a}f(x)\Big]^n.

Direct substitution: for any polynomial, and for any rational/radical function where the denominator is nonzero and the radicand is valid, lim⁡x→af(x)=f(a)\displaystyle\lim_{x\to a}f(x)=f(a).

Example
Factoring away a 00\tfrac{0}{0} form

Evaluate lim⁡x→3x2−9x−3\displaystyle\lim_{x\to 3}\frac{x^2-9}{x-3}.

Direct substitution gives 9−93−3=00\dfrac{9-9}{3-3}=\dfrac{0}{0} (indeterminate), so we factor and cancel:

x2−9x−3=(x−3)(x+3)x−3=x+3(x≠3).\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3\quad (x\ne 3).

Now substitute: lim⁡x→3(x+3)=3+3=6\displaystyle\lim_{x\to 3}(x+3)=3+3=6.

Example
Rationalizing a 00\tfrac{0}{0} form

Evaluate lim⁡x→0x+4−2x\displaystyle\lim_{x\to 0}\frac{\sqrt{x+4}-2}{x}. Substitution gives 00\tfrac{0}{0}. Multiply by the conjugate:

x+4−2x⋅x+4+2x+4+2=(x+4)−4x(x+4+2)=xx(x+4+2)=1x+4+2.\frac{\sqrt{x+4}-2}{x}\cdot\frac{\sqrt{x+4}+2}{\sqrt{x+4}+2} =\frac{(x+4)-4}{x\big(\sqrt{x+4}+2\big)} =\frac{x}{x\big(\sqrt{x+4}+2\big)}=\frac{1}{\sqrt{x+4}+2}.

Then lim⁡x→01x+4+2=12+2=14.\displaystyle\lim_{x\to 0}\frac{1}{\sqrt{x+4}+2}=\frac{1}{2+2}=\frac{1}{4}.

Tip

Tip: Always try direct substitution first. Only if it produces the indeterminate form 00\tfrac{0}{0} do you switch tools: factor and cancel (for polynomial ratios) or multiply by the conjugate (when a square root is involved).

One-Sided Limits; When a Limit Fails; Infinite Limits

Concept
One-sided limits and existence

The right-hand limit lim⁡x→a+f(x)\displaystyle\lim_{x\to a^{+}}f(x) uses inputs slightly larger than aa; the left-hand limit lim⁡x→a−f(x)\displaystyle\lim_{x\to a^{-}}f(x) uses inputs slightly smaller. The two-sided limit exists only when they agree:

lim⁡x→af(x)=L  ⟺  lim⁡x→a−f(x)=lim⁡x→a+f(x)=L.\lim_{x\to a}f(x)=L \iff \lim_{x\to a^{-}}f(x)=\lim_{x\to a^{+}}f(x)=L.

A limit fails to exist (DNE) when: the one-sided limits differ (a jump), the function grows without bound (infinite), or the function oscillates wildly near aa.

Example
A limit that fails to exist

For the graph above, lim⁡x→2−f(x)=2\displaystyle\lim_{x\to 2^{-}}f(x)=2 but lim⁡x→2+f(x)=4\displaystyle\lim_{x\to 2^{+}}f(x)=4. Since the one-sided limits disagree, lim⁡x→2f(x)\displaystyle\lim_{x\to 2}f(x) does not exist. (Note f(2)=4f(2)=4 from the closed dot---but that has no bearing on the two-sided limit.)

Concept
Infinite limits and vertical asymptotes

If f(x)f(x) increases or decreases without bound as x→ax\to a, we write lim⁡x→af(x)=∞\displaystyle\lim_{x\to a}f(x)=\infty (or −∞-\infty). This is a way of describing a limit that fails to exist; the line x=ax=a is a vertical asymptote. For f(x)=1xf(x)=\dfrac1x:

lim⁡x→0−1x=−∞,lim⁡x→0+1x=+∞.\lim_{x\to 0^{-}}\frac1x=-\infty,\qquad \lim_{x\to 0^{+}}\frac1x=+\infty.
Tip

Tip: To find a vertical asymptote of a rational function, look for inputs that make the denominator zero but the numerator nonzero after all common factors are cancelled. A factor that cancels gives a hole, not an asymptote.

Limits at Infinity and End Behavior

Concept
What happens as x→±∞x\to\pm\infty

lim⁡x→∞f(x)=L\displaystyle\lim_{x\to\infty}f(x)=L means the outputs level off at the height y=Ly=L as xx grows large; the line y=Ly=L is a horizontal asymptote. Two building blocks:

lim⁡x→±∞1xn=0 (n>0),lim⁡x→∞xn=∞ (n>0).\lim_{x\to\pm\infty}\frac{1}{x^n}=0\ (n>0),\qquad \lim_{x\to\infty}x^n=\infty\ (n>0).

Rational functions p(x)q(x)\dfrac{p(x)}{q(x)}---compare degrees:

  • [leftmargin=5mm]
  • degree top << degree bottom ⇒\Rightarrow limit 00 (H.A. y=0y=0);
  • degree top == degree bottom ⇒\Rightarrow limit == ratio of leading coefficients;
  • degree top >> degree bottom ⇒\Rightarrow limit ±∞\pm\infty (no H.A.).
Example
A limit at infinity

Evaluate lim⁡x→∞2x2−xx2+1\displaystyle\lim_{x\to\infty}\frac{2x^2-x}{x^2+1}. Divide numerator and denominator by the highest power, x2x^2:

2x2−xx2+1=2−1x1+1x2 ⟶ 2−01+0=2.\frac{2x^2-x}{x^2+1}=\frac{2-\tfrac1x}{1+\tfrac{1}{x^2}}\ \longrightarrow\ \frac{2-0}{1+0}=2.

The degrees match, so the limit is the ratio of leading coefficients, 21=2\tfrac21=2. The line y=2y=2 is a horizontal asymptote.

Tip

Tip: For polynomials, end behavior is governed entirely by the leading term: lim⁡x→±∞(3x3−⋯ )\lim\limits_{x\to\pm\infty}(3x^3-\cdots) behaves like 3x33x^3. Lower-degree terms are negligible for large ∣x∣|x|.

Continuity at a Point; Types of Discontinuity

Concept
The three-part continuity test

A function ff is continuous at x=ax=a if all three hold:

(1) f(a) is defined,(2) lim⁡x→af(x) exists,(3) lim⁡x→af(x)=f(a).\text{(1)}\ f(a)\ \text{is defined},\quad \text{(2)}\ \lim_{x\to a}f(x)\ \text{exists},\quad \text{(3)}\ \lim_{x\to a}f(x)=f(a).

Informally: you can draw the graph through x=ax=a without lifting your pencil. If any part fails, ff has a discontinuity there.

Concept
Classifying discontinuities
  • [leftmargin=5mm]
  • Removable (a hole): lim⁡x→af\lim\limits_{x\to a}f exists but ≠f(a)\ne f(a) (or f(a)f(a) is undefined). “Removable” because redefining one point would fix it.
  • Jump: the left- and right-hand limits both exist but disagree.
  • Infinite: the function blows up to ±∞\pm\infty (a vertical asymptote).

% Removable

  % Jump

  % Infinite

Example
Testing continuity

Is f(x)=x2−9x−3f(x)=\dfrac{x^2-9}{x-3} continuous at x=3x=3? Here f(3)=00f(3)=\tfrac00 is undefined, so part (1) fails---ff is discontinuous at 33. Because lim⁡x→3f(x)=6\lim\limits_{x\to3}f(x)=6 exists, the discontinuity is removable: defining f(3)=6f(3)=6 would patch the hole.

Tip

Tip: To classify a discontinuity from an equation, factor the rational expression. A factor that cancels signals a removable hole; a leftover zero in the denominator signals an infinite discontinuity (vertical asymptote).

The Tangent-Line Problem, the Derivative, and the Area Problem

Concept
From secant slopes to the tangent slope

The slope of the secant line through (x,f(x))\big(x,f(x)\big) and (x+h,f(x+h))\big(x+h,f(x+h)\big) is the difference quotient f(x+h)−f(x)h\dfrac{f(x+h)-f(x)}{h}. Letting h→0h\to 0 slides the second point into the first; the secant slopes approach the slope of the tangent line. That limiting slope is the derivative:

f′(x)=lim⁡h→0f(x+h)−f(x)h.f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.
Example
Derivative from the limit definition

Find f′(x)f'(x) for f(x)=x2f(x)=x^2.

f(x+h)−f(x)h=(x+h)2−x2h=x2+2xh+h2−x2h=2xh+h2h=2x+h.\frac{f(x+h)-f(x)}{h}=\frac{(x+h)^2-x^2}{h}=\frac{x^2+2xh+h^2-x^2}{h}=\frac{2xh+h^2}{h}=2x+h.

Then

f′(x)=lim⁡h→0(2x+h)=2x.f'(x)=\lim_{h\to 0}(2x+h)=2x.

So the tangent slope at x=3x=3 is f′(3)=6f'(3)=6.

Concept
The area problem

The other great problem of calculus is finding the area under a curve y=f(x)y=f(x) over [a,b][a,b]. Approximate it with nn thin rectangles of width Δx=b−an\Delta x=\tfrac{b-a}{n}, sum their areas, and take a limit:

Area=lim⁡n→∞∑k=1nf(xk) Δx.\text{Area}=\lim_{n\to\infty}\sum_{k=1}^{n} f(x_k)\,\Delta x.

As n→∞n\to\infty the rectangles shrink and the approximation becomes exact. This limit is the definite integral.

Tip

Tip: Both headline ideas of calculus are limits: the derivative is a limit of secant slopes (h→0h\to0), and the definite integral is a limit of rectangle-sum areas (n→∞n\to\infty). Mastering limits now is what makes calculus possible later.

Going Deeper: Advanced Limit Ideas

Concept
A bigger toolbox for the 00\tfrac{0}{0} form

When direct substitution gives 00\tfrac{0}{0}, the expression hides a common factor you must expose and cancel. Choose the tool by what you see:

  • [leftmargin=5mm]
  • Polynomial ratio ⇒\Rightarrow factor top and bottom, cancel the shared factor.
  • A square root ⇒\Rightarrow multiply by the conjugate to turn  \sqrt{\ } differences into plain differences.
  • A complex fraction (fractions stacked inside a fraction) ⇒\Rightarrow combine the small fractions over a common denominator first, then simplify the big fraction.
  • A trig ratio ⇒\Rightarrow rewrite in terms of sin⁡\sin and cos⁡\cos and steer toward the small-angle facts lim⁡x→0sin⁡xx=1\displaystyle\lim_{x\to0}\frac{\sin x}{x}=1 and lim⁡x→01−cos⁡xx2=12\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12.
Example
A complex-fraction limit

Evaluate lim⁡x→01x+2−12x\displaystyle\lim_{x\to 0}\frac{\dfrac{1}{x+2}-\dfrac{1}{2}}{x}. Substitution gives 00\tfrac{0}{0}. First combine the top over the common denominator 2(x+2)2(x+2):

1x+2−12=2−(x+2)2(x+2)=−x2(x+2).\frac{1}{x+2}-\frac12=\frac{2-(x+2)}{2(x+2)}=\frac{-x}{2(x+2)}.

Dividing by xx means multiplying by 1x\tfrac1x, and the xx's cancel:

1x⋅−x2(x+2)=−12(x+2) ⟶ −12(0+2)=−14.\frac{1}{x}\cdot\frac{-x}{2(x+2)}=\frac{-1}{2(x+2)}\ \longrightarrow\ \frac{-1}{2(0+2)}=-\frac14.
Example
A three-factor trig limit: lim⁡x→0tan⁡x−sin⁡xx3\lim_{x\to0}\frac{\tan x-\sin x}{x^3}

Substitution gives 00\tfrac00. Rewrite tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x} and factor sin⁡x\sin x out of the numerator:

tan⁡x−sin⁡x=sin⁡xcos⁡x−sin⁡x=sin⁡x ⁣(1−cos⁡xcos⁡x).\tan x-\sin x=\frac{\sin x}{\cos x}-\sin x=\sin x\!\left(\frac{1-\cos x}{\cos x}\right).

Now split the x3x^3 in the denominator as x⋅x2x\cdot x^2 and group into three familiar pieces:

tan⁡x−sin⁡xx3=sin⁡x (1−cos⁡x)x3cos⁡x=sin⁡xx⏟→ 1⋅1−cos⁡xx2⏟→ 12⋅1cos⁡x⏟→ 1.\begin{aligned} \frac{\tan x-\sin x}{x^3} &=\frac{\sin x\,(1-\cos x)}{x^3\cos x} =\underbrace{\frac{\sin x}{x}}_{\to\,1}\cdot \underbrace{\frac{1-\cos x}{x^2}}_{\to\,\frac12}\cdot \underbrace{\frac{1}{\cos x}}_{\to\,1}. \end{aligned}

Each factor has a known limit as x→0x\to0, so the product is 1⋅12⋅1=121\cdot\tfrac12\cdot1=\boxed{\dfrac12}.

Concept
Limits of the floor function ⌊x⌋\lfloor x\rfloor

The floor ⌊x⌋\lfloor x\rfloor is the greatest integer ≤x\le x; its graph is a staircase of flat steps with jumps at every integer. So at an integer nn the one-sided limits disagree:

lim⁡x→n−⌊x⌋=n−1,lim⁡x→n+⌊x⌋=n⇒lim⁡x→n⌊x⌋ DNE.\lim_{x\to n^{-}}\lfloor x\rfloor=n-1,\qquad \lim_{x\to n^{+}}\lfloor x\rfloor=n\quad\Rightarrow\quad \lim_{x\to n}\lfloor x\rfloor\ \text{DNE}.

Between integers the floor is constant, so for any non-integer cc, lim⁡x→c⌊x⌋=⌊c⌋\displaystyle\lim_{x\to c}\lfloor x\rfloor=\lfloor c\rfloor (continuous there).

Concept
Piecewise functions: solving for TWO parameters

Suppose a piecewise ff has two unknown constants a,ba,b and you want it smooth (continuous and differentiable) at the split point x=cx=c. You get two equations from two conditions:

(continuity)lim⁡x→c−f=lim⁡x→c+f,(differentiability)lim⁡x→c−f′=lim⁡x→c+f′.\text{(continuity)}\quad \lim_{x\to c^{-}}f=\lim_{x\to c^{+}}f,\qquad \text{(differentiability)}\quad \lim_{x\to c^{-}}f'=\lim_{x\to c^{+}}f'.

Match the heights for continuity and the slopes for differentiability, then solve the resulting 2×22\times 2 system for aa and bb.

Example
Making a piecewise function differentiable

Find a,ba,b so that

f(x)={x2,x≤1,ax+b,x>1,f(x)=\begin{cases} x^2, & x\le 1,\\[2pt] ax+b, & x>1,\end{cases}

is differentiable at x=1x=1.

Slopes (differentiability). For x<1x<1, f′(x)=2x→2f'(x)=2x\to 2; for x>1x>1, f′(x)=af'(x)=a. Matching slopes gives a=2a=2.

Heights (continuity). The left value is 12=11^2=1; the right value is a(1)+b=2+ba(1)+b=2+b. Matching gives 2+b=12+b=1, so b=−1b=-1.

Thus a=2, b=−1a=2,\ b=-1: the parabola hands off to the tangent line y=2x−1y=2x-1 with no corner.

Example
The derivative of a harder function from the definition

Find f′(x)f'(x) for f(x)=1xf(x)=\dfrac1x. Form the difference quotient and combine the small fractions:

f(x+h)−f(x)h=1x+h−1xh=x−(x+h)x(x+h)h=−hh x(x+h)=−1x(x+h).\frac{f(x+h)-f(x)}{h}=\frac{\dfrac{1}{x+h}-\dfrac1x}{h} =\frac{\dfrac{x-(x+h)}{x(x+h)}}{h} =\frac{-h}{h\,x(x+h)}=\frac{-1}{x(x+h)}.

The hh cancels, so

f′(x)=lim⁡h→0−1x(x+h)=−1x⋅x=−1x2.f'(x)=\lim_{h\to 0}\frac{-1}{x(x+h)}=\frac{-1}{x\cdot x}=-\frac{1}{x^2}.
Example
A limit at infinity with a radical

Evaluate lim⁡x→∞(x2+3x−x)\displaystyle\lim_{x\to\infty}\big(\sqrt{x^2+3x}-x\big). This is an ∞−∞\infty-\infty form, so multiply by the conjugate:

(x2+3x−x)⋅x2+3x+xx2+3x+x=(x2+3x)−x2x2+3x+x=3xx2+3x+x.\big(\sqrt{x^2+3x}-x\big)\cdot\frac{\sqrt{x^2+3x}+x}{\sqrt{x^2+3x}+x} =\frac{(x^2+3x)-x^2}{\sqrt{x^2+3x}+x} =\frac{3x}{\sqrt{x^2+3x}+x}.

Divide top and bottom by xx (and, since x>0x>0, note x2+3x=x1+3/x\sqrt{x^2+3x}=x\sqrt{1+3/x}):

3xx2+3x+x=31+3x+1 ⟶ 31+1=32.\frac{3x}{\sqrt{x^2+3x}+x}=\frac{3}{\sqrt{1+\tfrac3x}+1}\ \longrightarrow\ \frac{3}{\sqrt{1}+1}=\frac32.
Tip

Preview---area as a limit of Riemann sums. To find the exact area under y=x2y=x^2 on [0,1][0,1], cut it into nn rectangles of width Δx=1n\Delta x=\tfrac1n using right endpoints xk=knx_k=\tfrac{k}{n}:

Sn=∑k=1n(kn)2 ⁣⋅1n=1n3∑k=1nk2=1n3⋅n(n+1)(2n+1)6.S_n=\sum_{k=1}^{n}\Big(\tfrac{k}{n}\Big)^2\!\cdot\tfrac1n=\frac{1}{n^3}\sum_{k=1}^{n}k^2=\frac{1}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}.

As n→∞n\to\infty this tends to 26=13\dfrac26=\dfrac13, so the area is 13\tfrac13. Every definite integral is exactly this kind of limit.

Formulas, Proofs & Tips

Tip
The difference quotient and the derivative
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}

What it means. The slope of the tangent line — the instantaneous rate of change.

Example. For f(x)=x2f(x)=x^2: f′(x)=lim⁡h→0(x+h)2−x2h=2xf'(x)=\lim_{h\to0}\tfrac{(x+h)^2-x^2}{h}=2x.

Why it works. f(x+h)−f(x)h\tfrac{f(x+h)-f(x)}{h} is the slope of the secant through two nearby points. Letting h→0h\to 0 slides the second point into the first, so the secant becomes the tangent.

Tip. The hh must cancel before you substitute h=0h=0; otherwise you get the meaningless 00\tfrac00.

Tip
L'Hôpital's Rule
lim⁡x→cf(x)g(x)=lim⁡x→cf′(x)g′(x)when 00 or ∞∞\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)} \quad\text{when } \tfrac{0}{0} \text{ or } \tfrac{\infty}{\infty}

What it means. If a limit gives the indeterminate form 00\tfrac{0}{0} or ∞∞\tfrac{\infty}{\infty}, differentiate the top and bottom separately and try again.

Why it works. Near cc both functions are approximated by their tangent lines, f(x)≈f′(c)(x−c)f(x)\approx f'(c)(x-c) and g(x)≈g′(c)(x−c)g(x)\approx g'(c)(x-c). The common factor (x−c)(x-c) cancels in the ratio, leaving f′(c)g′(c)\tfrac{f'(c)}{g'(c)}.

Tip. Check the form FIRST — applying it to a limit that isn't indeterminate gives a wrong answer. It is not the quotient rule: do not differentiate fg\tfrac{f}{g} as a whole.

Tip
The Squeeze Theorem
g(x)≤f(x)≤h(x)  and lim⁡g=lim⁡h=L ⟹ lim⁡f=Lg(x)\le f(x)\le h(x) \ \text{ and } \lim g=\lim h=L \ \Longrightarrow\ \lim f = L

What it means. If a function is trapped between two others that meet at LL, it must go to LL too.

Example. Since −x2≤x2sin⁡1x≤x2-x^2\le x^2\sin\tfrac1x\le x^2 and both bounds →0\to0, the limit at 00 is 00.

Why it works. Beyond some point ff never leaves the gap between gg and hh. As that gap closes to zero width around LL, ff has nowhere else to go.

Tip. The classic use is lim⁡x→0x2sin⁡1x=0\lim_{x\to 0}x^{2}\sin\tfrac1x = 0, squeezed by −x2-x^{2} and x2x^{2}.