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Could someone please explain what a limit is?

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  1. May 25, 2015 #1
    I've been searching around trying to understand them. About to take calculus and I want to be prepared. Could someone explain what they are and give a few typical limit problems and solve them

    Thank you
     
    Last edited: May 25, 2015
  2. jcsd
  3. May 25, 2015 #2

    PeroK

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  4. May 25, 2015 #3

    Mark44

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    A limit can be used to tell you the behavior of a function near some point. It's most useful for functions that aren't defined at the point in question.

    Some limits are very easy to calculate, either because the function in question is continuous at the point. For example, ##\lim_{x \to 2} 3x = 6##. Here the function is f(x) = 3x. Like all polynomials, this function is continuous everywhere, so taking the limit is as simple as evaluating the function at x = 2.

    More interesting are limits where the function is not continuous at a particular point. Even so, it's possible for the limit to exist. Here's an example.
    ##\lim_{x \to 1}\frac{x^2 - 1}{x - 1}##. It turns out that this limit exists and is equal to 2. If you graph ##y = \frac{x^2 - 1}{x - 1}##, its graph is identical to the graph of y = x + 1, except that there is a "hole" at the point (1, 2). Taking the limit gives us the y-coordinate at the hole.

    Probably the most important use of the limit in calculus is that it is used in the definition of the derivative of a function at a particular point. It is also used in the definition of one type of integral.
     
  5. May 25, 2015 #4

    Ssnow

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    I want to add another example that you can joint in yours collection, for ##x\rightarrow +\infty##

    [tex]\lim_{x\rightarrow +\infty} \frac{x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x}{x^2+x}=\frac{\infty}{\infty}[/tex]

    we call ##\frac{\infty}{\infty},\frac{0}{0},0\cdot \infty, \left(0^{+}\right)^{0^{+}},+\infty-\infty##
    indecision forms and we cannot say what is the exact value . In order to eliminate it we can use some tricks as:

    [tex]\lim_{x\rightarrow +\infty} \frac{x^9\left[1+x^{-1}+x^{-2}+x^{-3}+x^{-4}+x^{-5}+x^{-6}+x^{-7}+x^{-8}\right]}{x^{2}\left[1+x^{-1}\right]}\sim \lim_{x\rightarrow +\infty}\frac{x^{9}}{x^{2}}\sim \lim_{x\rightarrow +\infty}x^{7}=+\infty[/tex]
    limits can be ##\infty##, we include this symbol in ##\mathbb{R}## and we call ##\overline{\mathbb{R}}=\mathbb{R}\cup\{\infty\}## the extension with a ''partial arithmetization'' of the ##\infty## with rules ## \pm\infty \pm\infty=\pm\infty, \frac{c}{\infty}=0, \frac{c}{0^{+}}=\infty## with ##c>0## and ##-\infty## with ##c<0##. Obviously ##(\pm\infty)\cdot(\pm\infty)=+\infty## .... Remember important concepts

    1)if limit exists then it is unique;
    2)''in general'' algebraic operations are according with the the limit operator (so the limit of a sum is the sum of the limits ... );
    3)in the limit operation we are interested to understand what happen to a function ''near'' to the point ##x_{0}## (if ##x_{0}=+\infty## we consider the set of '' far'' points as a neighborhood of ##\infty##).
     
  6. May 25, 2015 #5

    Mark44

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    No.
    We never write a limit as ##\frac{\infty}{\infty}##.
    They are called indeterminate forms, not indecision forms.
     
  7. May 26, 2015 #6

    Ssnow

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    Yah, sorry in Italy we usual call the ''indeterminate forms'' as '' indeterminate or indecision forms'', thank you Mark44 for the clarification ...
     
  8. May 26, 2015 #7

    phion

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    L-Hopital's rule is awesome.
     
  9. May 26, 2015 #8

    Mark44

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    But it's not without its problems. For some problems, using L'Hopital's Rule just gets you back to the same indeterminate form you started with, so is no help, but simpler, non-calculus methods (such as removing common factors from numerator and denominator or multiplying by the conjugate) might be helpful.

    That's not to say that L'Hopital's Rule isn't helpful, but it's important to have more than one tool in your kit.
     
  10. May 26, 2015 #9

    Ssnow

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    Mark44 has reason, for example this limit

    [tex] \lim_{x\rightarrow +\infty}\frac{\sqrt{1+x^2}}{x} [/tex]

    is a ''perpetual motion'' limit because after two times that you apply De L'Hopital you obtain the same form ....
     
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