Latest revision as of 23:46, 6 December 2015
Limits
$\lim\limits_{x \to a} f(x) = L$
Definition
the limit of $f(x)$ as $x \to a$ is $L$:
- if $\forall \varepsilon > 0 \ \ \exists\, \delta > 0$ s.t. $x \ne a$ is within $\delta$ of $a$
- then $f(x)$ is within $\varepsilon$ of $L$
Interpretation
Interpretation:
- if as $x$ gets closer to $a$, $f(x)$ gets closer to $L$
- then $L$ is the limit of $f(x)$
or
- choose some output tolerance $\varepsilon$
- then there exists some input tolerance $\delta$ such that $L$ lies within $\varepsilon$ of $f(x)$
- if you change $\varepsilon$, you need to be able to update $\delta$ - this has to be true for any $\varepsilon$
$x \to \infty$
- $a = \infty$ then
- $\lim\limits_{x \to \infty} f(x) = L$
What does it mean? Can think if it as of the "end" of the real line
Definition:
- the definition needs to be slightly adapted
- if $\forall \varepsilon > 0 \ \ \exists\, M > 0$ s.t. $|f(x) - L| < \varepsilon$ when $x > M$
- if there's no such $M$, then the limit does not exist
Interpretation:
- instead of input tolerance $\infty \pm \delta, we mean some very large $M$:
- we always can find some large $M$ after which $f(x)$ is always within $\varepsilon$ of $L$
as $\varepsilon$ becomes tighter, we still should be able to find larger $M$ for which the function lies within $\varepsilon$
Non-Existent Limits
Some functions are not well-behaved, so things can go wrong
Discontinuity
The limit does not exist because the limit from the left and the limit from the right are not equal.
Blow-Up
The function has a vertical asymptote:
- functions gets to $\infty$ as $x \to a$
Oscillation
The function oscillates up and down as the input approaches some value
Rules
Rules for limits
Suppose $\lim\limits_{x \to a} f(x)$ and $\lim\limits_{x \to a} g(x)$ exist. Then
- Sum Rule: $\lim \big(f(x) + g(x) \big) = \lim f(x) + \lim g(x)$
- Product Rule: $\lim \big(f(x) \cdot g(x) \big) = \left[\lim f(x)\right] \cdot \left[ \lim g(x) \right]$
- Quotient Rule: $\lim \cfrac{f(x)}{g(x)} = \cfrac{\lim f(x)}{\lim g(x)}$ (when $\lim g(x) \ne 0$)
- Chain Rule (or Composition Rule): $\lim f \big(g(x) \big) = f \big(\lim g(x) \big)$ if $f$ is continuous
$$\lim_{x \to a} \cfrac{f(x)}{g(x)} = \lim_{x \to a} \cfrac{f'(x)}{g'(x)}$$
Gives a way to solve ambiguous limits:
- $\lim \cfrac{0}{0}$
- $\lim \cfrac{\infty}{\infty}$
- $\lim \infty \cdot 0$
- $\lim (\infty - \infty)$
- $\lim \infty^0$
Some Important Limits
$\lim\limits_{x \to 0} \cfrac{\sin x}{x}$
What is $\lim\limits_{x \to 0} \cfrac{\sin x}{x}$?
- cannot apply the Quotient Rule because $x \to 0$
- let's Taylor Expand $\sin x$
- $\lim\limits_{x \to 0} \cfrac{\sin x}{x} = \lim\limits_{x \to 0} \cfrac{x - \frac{1}{3!}\, x^3 + \frac{1}{5!}\, x^5 - \ ...}{x} = \ ...$
- $... \ = \lim\limits_{x \to 0} \cfrac{x\, \left(1 - \frac{1}{3!}\, x^2 + \frac{1}{5!}\, x^4 - \ ... \ \right)}{x} = \ ...$
- $... \ = \lim\limits_{x \to 0} \left(1 - \frac{1}{3!}\, x^2 + \frac{1}{5!}\, x^4 + \ ... \ \right) = 1$
Can also show this using the L'Hopital's Rule
- $\lim\limits_{x \to 0} \cfrac{\sin x}{x} = \lim\limits_{x \to 0} \cfrac{\cos x}{1} = 1$
Sources