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Transcript of Limits
Now we use |f(x)-L|<ε. From plugging 4 into 2x-1, we know L is equal to 7. Therefore, we get
Now, we need to find a relationship between δ and ε. To do that, we will need to get |x-4| to look like |(2x-1)-7|. Fortunately, in this case its very simple; all we have to do is multiply |x-4| by 2, giving us
Now, we can establish a direct relationship between the two, namely,
Therefore, our L must have been correct, which means as x gets closer and closer to 4, f(x) will get closer and closer to 7; the limit of 2x-1 as x approaches 4 is 7, or, lim┬(x→4)2x-1=7 This is the easy part. Most of the time, to evaluate a limit,
all you have to do is plug it in.
For example, lets use our earlier limit, the limit of 2x-1 as x approaches 4.
Plugging in, we get the limit as 7, which is confirmed by the graph.
Sometimes, it will be a little more complicated than that, but we'll deal with that later Ok, so now that we know what a limit is and how to take one, why do we care? Whats the point of limits? Well, they at least three applications; asymptotoes, holes in a function, and derivatives An asymptote is an imaginary line on a graph, which deals with infinite. A vertical asymptote is an imaginary line of the form x=c, where c is a constant, such that as x gets closer to c, f(x) becomes infinite. A horizontal aysmptote is a line (usually) of the form y=c, such that as x becomes infinite, f(x) approaches c Limits are useful for calculating horizontal asypmtotes. To find the horizontal asymptote, simply take the limit of the function as x approaches infinite. But be careful, the horizontal asymptote is not always horizontal. When a function has a hole in it, it has no
actual value at that point. However, we
can solve for what the function should be
at that point, by taking the limit of the
function as x approaches that value. Derivatives are probably the most important part of calculus. A derivative can be written as a limit, using the difference quotient; lim(h→0)(f(x+h)-f(x))/h
An integral is the opposite of a derivative. L'Hospital's rule deals with finding the limit of the quotient of two functions, when both of the functions are zero or infinite. The rule states, that the limit of the two functions at that point will be the same as the limit of the derivative of the top divided by the derivative of the bottom. Lets take something we already know, say, lim(x→0)sinx/x. We know from previous experiance that this is equal to one, but we can confirm this by L'Hospital. Taking the derivative of the top and the bottom leaves lim(x→0)cosx/1, which simplifies to 1/1, or 1. Now lets try the other form, infinite over infinite. Take
lim(x→∞)e^x/lnx. By L'Hospital, this becomes lim(x→∞)e^x/(1/x), which is infinite. The function f(x)=(x-2)/(x-3) has both a vertical and a horizontal asymptote. The vertical asymptote occurs at x=3, and the horizontal asymptote occurs at y=1 Lets say we were graphing f(x)=((2x-1)(x-4))/(x-4). Then, the graph would have a hole in it at x=4. Howeverr, if we take the limit as x approaches 4, we will get what the value the graph should have, of 7