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

$\frac{0}{0}$

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.

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

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]- $lim_{t \to 0} (\frac{1}{t \sqrt{t + 1}} - \frac{1}{t})$
if $t = 0$ , 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 $\frac{1}{2}$ .

Limit at infinity

Forcefully factorize dominant term
Rationalize
Combine

Contributors

Ajitani Hifumi

Changelog

Last edited 9 days ago

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Limit ​

Sidedness ​

Properties & theorems ​

Direct substitution property ​

Evaluation ​

Indeterminate forms ​

00 ​

∞∞ ​

∞−∞ ​

Limit at infinity ​