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Limits and continuity

The limit describes which value a function approaches as the variable approaches a point, even if the function is not defined right there. It is the basis of continuity, the derivative and the integral. This topic explains the idea, one-sided limits, how to resolve indeterminate forms and limits at infinity.

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The idea of a limit

We say that $lim_{x \to a} f (x) = L$ if the values of $f (x)$ stay as close to $L$ as we want by taking $x$ close enough to $a$ (but different from $a$ ). What happens at $x = a$ does not matter: the function may not exist there or may have another value.

f (x) = (x^{2} - 1) / (x - 1)

equals

x + 1

except at

x = 1

, where it has a hole. Its limit at 1 is 2.

One-sided limits look at one side only: $x \to a^{-}$ (values less than $a$ ) and $x \to a^{+}$ (greater). The limit exists if and only if both one-sided limits exist and are equal.

∣ x ∣/ x

- 1

to the left of 0 and

1

to the right: the one-sided limits do not agree.

A function is continuous at $a$ if $f (a)$ exists and equals the limit: $lim_{x \to a} f (x) = f (a)$ . Polynomials, exponentials, sine and cosine are continuous everywhere; quotients are where the denominator is not zero; roots and logarithms, within their domain.

How to compute it

1. Direct substitution. If the function is continuous at $a$ , the limit is $f (a)$ . It is the first thing to try.

2. Indeterminate forms. If substitution gives a form such as $0/0$ or $\infty/\infty$ , the value cannot be read directly: the expression must be transformed.

Form	What to do
$0/0$ con polinomios,with polynomials	factor and cancel the common factor $(x - a)$
$0/0$ con raíces,with roots	multiply by the conjugate
$0/0$ , $\infty/\infty$	L'Hôpital's rule: $lim \frac{f}{g} = lim \frac{f ^{'}}{g ^{'}}$
$0 \cdot \infty$	rewrite as a quotient $\frac{f}{1/ g}$
$\infty - \infty$	common denominator or conjugate
$1^{\infty}$ , $0^{0}$ , $\infty^{0}$	write $f^{g} = e^{g l n f}$ and compute the limit of the exponent

Two notable limits come up again and again:

x \to 0 lim \frac{sin x}{x} = 1 x \to \infty lim (1 + \frac{1}{x})^{x} = e

sin x / x

is not defined at 0, but its limit there is 1.

3. Limits at infinity. In a quotient of polynomials the highest-degree terms dominate: with equal degrees the limit is the ratio of the leading coefficients; if the numerator dominates it goes to $\pm \infty$ ; if the denominator does, to 0. If $f (x) \to L$ as $x \to \infty$ , the line $y = L$ is a horizontal asymptote; if $f (x) \to \pm \infty$ as $x \to a$ , the line $x = a$ is a vertical asymptote.

How the calculator does it: it tries direct substitution, checking that the exact value matches the numerical one; for indeterminate forms it factors, compares degrees or applies L'Hôpital, and at a finite point it always computes both one-sided limits. If none of that works, it gives a numerical estimate and says so.

Common mistakes

Saying that $0/0 = 0$ or $0/0 = 1$ : it is an indeterminate form, not a number.
Applying L'Hôpital without checking that the form is $0/0$ or $\infty/\infty$ (or differentiating the whole quotient instead of numerator and denominator separately).
Forgetting one-sided limits when there are absolute values, piecewise functions or a denominator that changes sign.
Thinking that $1^{\infty} = 1$ : $(1 + 1/ x)^{x}$ tends to $e \approx 2.718$ .

Worked examples

Example 1Factor and cancel

Compute

x \to 1 lim \frac{x ^{2} - 1}{x - 1}

Solution

Substituting $x = 1$ gives $\frac{0}{0}$ : indeterminate. Since $x = 1$ is a root of the numerator and of the denominator, both have the factor $(x - 1)$ .
$\frac{x ^{2} - 1}{x - 1} = \frac{( x - 1 ) ( x + 1 )}{x - 1} = x + 1$ for $x \neq = 1$ (the limit does not look at $x = 1$ , so canceling is allowed).
$x + 1$ is continuous: $lim_{x \to 1} (x + 1) = 2$ .

x \to 1 lim \frac{x ^{2} - 1}{x - 1} = 2

Solve it in Simuveliq

In Math → Symbolic calculator, topic Calculus, operation Limits: type (x^2-1)/(x-1), Tends to = 1 and press Compute. The Answer is 2; the Steps show each transformation.
The steps are the same: 0/0, factor with $(x - 1)$ and cancel. The small graph shows the line with the hole.

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Example 2A root in the numerator

Compute

x \to 0 lim \frac{x + 4 - 2}{x}

Solution

At $x = 0$ : $\frac{4 - 2}{0} = \frac{0}{0}$ . Multiply top and bottom by the conjugate $x + 4 + 2$ :
$\frac{( x + 4 - 2 ) ( x + 4 + 2 )}{x ( x + 4 + 2 )} = \frac{x + 4 - 4}{x ( x + 4 + 2 )} = \frac{1}{x + 4 + 2}$ .
Now substitute: $\frac{1}{2 + 2} = \frac{1}{4}$ . L'Hôpital gives the same: $lim \frac{1/ ( 2 x + 4 )}{1} = \frac{1}{4}$ .

\frac{1}{4}

Solve it in Simuveliq

In Math → Symbolic calculator, topic Calculus, operation Limits: type (sqrt(x+4)-2)/x, Tends to = 0 and press Compute. The Answer is 1/4; the Steps show each transformation.
The calculator solves it by L'Hôpital (the second route of the solution) and checks both one-sided limits.

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Example 3L'Hôpital twice

Compute

x \to 0 lim \frac{1 - cos x}{x ^{2}}

Solution

At $x = 0$ : $\frac{1 - 1}{0} = \frac{0}{0}$ . L'Hôpital: $lim \frac{sin x}{2 x}$ , still $\frac{0}{0}$ .
L'Hôpital again: $lim \frac{cos x}{2} = \frac{1}{2}$ . (Or with the notable limit: $\frac{sin x}{2 x} = \frac{1}{2} \cdot \frac{sin x}{x} \to \frac{1}{2}$ .)

\frac{1}{2}

Solve it in Simuveliq

In Math → Symbolic calculator, topic Calculus, operation Limits: type (1-cos(x))/x^2, Tends to = 0 and press Compute. The Answer is 1/2; the Steps show each transformation.

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Example 4Limit at infinity

Compute

x \to \infty lim \frac{2 x ^{2} + 1}{3 x ^{2} - x}

and say which asymptote the function has.

Solution

Substituting gives $\frac{\infty}{\infty}$ . Divide top and bottom by the highest power, $x^{2}$ : $\frac{2 + 1/ x ^{2}}{3 - 1/ x}$ .
$1/ x^{2} \to 0$ and $1/ x \to 0$ , so the limit is $\frac{2}{3}$ : the ratio of the leading coefficients (same degree).
The line $y = 2/3$ is a horizontal asymptote.

\frac{2}{3}

; horizontal asymptote

y = \frac{2}{3}

Solve it in Simuveliq

In Math → Symbolic calculator, topic Calculus, operation Limits: type (2x^2+1)/(3x^2-x), Tends to = inf and press Compute. The Answer is 2/3; the Steps show each transformation.

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Example 5When the one-sided limits disagree

Does

x \to 0 lim \frac{∣ x ∣}{x}

exist?

Solution

For $x < 0$ , $∣ x ∣ = - x$ and the function equals $- 1$ : $lim_{x \to 0^{-}} = - 1$ .
For $x > 0$ , $∣ x ∣ = x$ and it equals $1$ : $lim_{x \to 0^{+}} = 1$ .
The one-sided limits are different: the limit does not exist (the function jumps).

It does not exist:

- 1

from the left,

1

from the right.

Solve it in Simuveliq

In Math → Symbolic calculator, topic Calculus, operation Limits: type abs(x)/x, Tends to = 0 and press Compute. The Answer is «The limit does not exist: from the left it tends to -1 and from the right to 1»; the Steps show each transformation.
With Side = «Left (−)» (the next example in the list) it gives just the one-sided limit, $- 1$ .

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Example 6The form 1 to the infinity

Compute

x \to \infty lim (1 + \frac{1}{x})^{x}

Solution

The base tends to 1 and the exponent to $\infty$ : form $1^{\infty}$ . Write it as an exponential: $(1 + \frac{1}{x})^{x} = e^{x l n (1 + 1/ x)}$ .
The exponent is $\infty \cdot 0$ ; as a quotient, $\frac{ln ( 1 + 1/ x )}{1/ x}$ is $\frac{0}{0}$ . L'Hôpital: $\frac{\frac{- 1/ x ^{2}}{1 + 1/ x}}{- 1/ x ^{2}} = \frac{1}{1 + 1/ x} \to 1$ .
The limit is $e^{1} = e \approx 2.7182818$ .

e \approx 2.7182818

Solve it in Simuveliq

In Math → Symbolic calculator, topic Calculus, operation Limits: type (1+1/x)^x, Tends to = inf and press Compute. The Answer is $e \approx 2.7182818$ ; the Steps show each transformation.

Open in the simulator →

Limits and continuity

The idea of a limit

How to compute it

Common mistakes

Worked examples

Example 1Factor and cancel

Solution

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Example 2A root in the numerator

Solution

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Example 3L'Hôpital twice

Solution

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Example 4Limit at infinity

Solution

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Example 5When the one-sided limits disagree

Solution

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Example 6The form 1 to the infinity

Solution

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