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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 $\varepsilon$

limit-def.png


$x \to \infty$

  • $a = \infty$ 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 $L$

limit-def-infty.png

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.

limit-discontinuity.png


Blow-Up

The function has a vertical asymptote:

  • functions gets to $\infty$ as $x \to a$

limit-blowup.png


Oscillation

The function oscillates up and down as the input approaches some value

limit-oscillation.png


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


L'Hopital's Rule

$$\lim_{x \to a} \cfrac{f(x)}{g(x)} = \lim_{x \to a} \cfrac{f'(x)}{g'(x)}$$

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