Limit

Sidedness

By default, $x$ can approach from either side
For left-side ( $x < a$ ) / right-side ( $a < x$ ) limit, superscript $a$ with $-$ / $+$
Superscript the limit with $-$ / $+$ depending on whether it is approached from the

[!example]+ Left-side limit
$f (x)$ has the left-side limit $M$ as $x$ approaches $a$ :
$lim_{x \to a^{-}} f (x) = M$

[!example]+ Right-side limit
$f (x)$ has the left-side limit $R$ as $x$ approaches $a$ :
$lim_{x \to a^{+}} f (x) = R$

+ Existence of a limit theorem

$lim_{x \to a} f (x) = L ⟺ lim_{x \to a^{-}} f (x) = lim_{x \to a^{+}} f (x) = L$
Otherwise, we say the limit D.N.E. (does not exist)

The limit is approached but never reached

Properties & theorems

$lim_{x \to a} k = k$
$lim_{x \to a} x = a$
$lim_{x \to a} k f (x) = k lim_{x \to a} f (x) = k L$
$lim_{x \to a} (f @ g) (x) = lim_{x \to a} f (x) @ lim_{x \to a} g (x) = L @ M$ , where $@$ is any of addition, subtraction, multiplication or division (when $@$ = division, $M \neq 0$ )
$lim_{x \to a} f^{n} (x) = (lim_{x \to a} f (x))^{n} = L^{n}$
$lim_{x \to a} \frac{\sin g (x)}{g (x)} = 1$ , where $a \in R$ or $a = \infty$ and $g (a) = 0$

Direct substitution property

If $f (x)$ is a polynomial with $x = a$ in its domain, then $lim_{x \to a} f (x) = f (a)$

Evaluation

For a generic limit $lim_{x \to a} f (x)$ :

If evaluating a piecewise function at boundary of two pieces, the left and right limits need to be considered;
Substitute $a$ into $f (x)$ using the direct substitution property. If the result is an indeterminate form, follow the steps below.

Indeterminate forms

Multiple ways to solve
All indeterminate forms can be converted to use L'Hôpital's Rule.

$\frac{0}{0}$

For a limit $lim_{x \to a} \frac{P (x)}{Q (x)}$ where $P (x)$ and $Q (x)$ are polynomials such that $P (a) = Q (a) = 0$ ,

first rationalize the fraction.

Because it is always possible to factorize a polynomial equal to 0, we factorize $\frac{P (x)}{Q (x)}$ either:

to $\frac{(x - a) P^{'} (x)}{(x - a) Q^{'} (x)}$ and cancel $(x - a)$ , or
with long division.

Now that it no longer contains the factor that makes it 0, we can solve the limit as usual.

[!example]- $lim_{x \to 1} \frac{\sqrt{2 + x^{2}} - \sqrt{4 x - x^{2}}}{x^{2} - 1}$
When $x = 1$ , limit = $\frac{0}{0}$ .
We rationalize the numerator:
$\begin{aligned} lim_{x \to 1} \frac{\sqrt{2 + x^{2}} - \sqrt{4 x - x^{2}}}{x^{2} - 1} & = lim_{x \to 1} (\frac{\sqrt{2 + x^{2}} - \sqrt{4 x - x^{2}}}{x^{2} - 1} \cdot \frac{\sqrt{2 + x^{2}} + \sqrt{4 x - x^{2}}}{\sqrt{2 + x^{2}} + \sqrt{4 x - x^{2}}}) \\ = lim_{x \to 1} \frac{2 + x^{2} - (4 x - x^{2})}{(x^{2} - 1) (\sqrt{2 + x^{2}} + \sqrt{4 x - x^{2}})} \\ = lim_{x \to 1} \frac{2 x^{2} - 4 x + 2}{(x^{2} - 1) (\sqrt{2 + x^{2}} + \sqrt{4 x - x^{2}})} \\ = 2 lim_{x \to 1} \frac{(x - 1)^{2}}{(x - 1) (x + 1) (\sqrt{x^{2} + 2} + \sqrt{4 x - x^{2}})} \\ = 2 lim_{x \to 1} \frac{x - 1}{(x + 1) (\sqrt{x^{2} + 2} + \sqrt{4 x - x^{2}})} \\ = 0 \end{aligned}$

$\frac{\infty}{\infty}$

For a limit $lim_{x \to \infty} \frac{P (x)}{Q (x)}$ evaluating to $\pm \frac{\infty}{\infty}$ ,
there are three possible answers:

$0$ when $d e g (P) < d e g (Q)$
$\pm \infty$ when $d e g (P) > d e g (Q)$
$R_{*}$ when $d e g (P) = d e g (Q)$

99% of the time, forcefully factorize the dominant term.
Otherwise, rationalize.

[!example]- $lim_{x \to \infty} \frac{x \sqrt{x + 1} (1 - 2 \sqrt{x^{2} - 3})}{7 - 6 x + 4 x^{3}}$
When $x \to \infty$ , limit = $- \frac{\infty}{\infty}$ .
We factorize:
$lim_{x \to \infty} \frac{x \sqrt{x + 1} (1 - 2 \sqrt{x^{2} - 3})}{7 - 6 x + 4 x^{3}} = lim_{x \to \infty} \frac{x \sqrt{x} \sqrt{1 + \frac{1}{x}} (1 - 2 | x | \sqrt{1 - \frac{3}{x^{2}}})}{x^{3} (\frac{7}{x^{3}} - \frac{6}{x^{2}} + 4)}$
Since $x \to \infty$ , $| x | = x$ .
$\begin{aligned} lim_{x \to \infty} \frac{x \sqrt{x} \sqrt{1 + \frac{1}{x}} (1 - 2 x \sqrt{1 - \frac{3}{x^{2}}})}{x^{3} (\frac{7}{x^{3}} - \frac{6}{x^{2}} + 4)} & = lim_{x \to \infty} \frac{x^{2} \sqrt{x} \sqrt{1 + \frac{1}{x}} (\frac{1}{x} - 2 \sqrt{1 - \frac{3}{x^{2}}})}{x^{3} (\frac{7}{x^{3}} - \frac{6}{x^{2}} + 4)} \\ = \frac{\sqrt{1 + \frac{1}{x}} (\frac{1}{x} - 2 \sqrt{1 - \frac{3}{x^{2}}})}{\sqrt{x} (\frac{7}{x^{3}} - \frac{6}{x^{2}} + 4)} \\ = 0 \end{aligned}$

$\infty - \infty$

Factorize
Combine
Rationalize

[!example]- $lim_{t \to 0} (\frac{1}{t \sqrt{t + 1}} - \frac{1}{t})$
When $t = 0$ , limit = $\frac{1}{0} - \frac{1}{0}$ .
Combining the two fractions gives $lim_{t \to 0} \frac{1 - \sqrt{t + 1}}{t \sqrt{t + 1}}$ .
$\begin{aligned} \frac{1 - \sqrt{t + 1}}{t \sqrt{t + 1}} & = \frac{t}{t \sqrt{t + 1} (1 + \sqrt{t + 1})} \\ = \frac{1}{\sqrt{t + 1} + t + 1} \end{aligned}$
Substituting $0$ into $t$ gives $\frac{1}{2}$ .

Others

Use L'Hôpital's Rule

Limit at infinity

Forcefully factorize dominant term
Rationalize
Combine

Contributors

Ajitani Hifumi

Changelog

Last edited about 1 month ago

View full history

Limit ​

Sidedness ​

Properties & theorems ​

Direct substitution property ​

Evaluation ​

Indeterminate forms ​

00 ​

∞∞ ​

∞−∞ ​

Others ​

Limit at infinity ​