Calculus, all content (2017 edition)
- One-sided limits from graphs
- One-sided limits from graphs: asymptote
- One-sided limits from graphs
- One-sided limits from tables
- 1-sided vs. 2-sided limits (graphical)
- Limits of piecewise functions: absolute value
- Connecting limits and graphical behavior (more examples)
Sal finds the one-sided limit of x²/(1-cosx) as x approaches zero from the right, using a table of values. Created by Sal Khan.
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- At1:20you verify that you're in radian mode. Why do you need to be in radian mode versus degrees?(69 votes)
- I think what is being asked is how can we tell from the question that we need to use radians. We could do this by checking the other values in the function and we would clearly see that the measurements are not in degrees. I do not see anywhere that the problem clearly states that the solution needs to be in radians, however the information used to provide the solution makes it clear we are talking about radians.(12 votes)
- okay what happens if approaching from both sides we dont get the same value ?(14 votes)
- One of the problems given to me during this section is estimating the limit as x approaches "f(x)=sin(x)ln(x)"
I am not sure what ln(x) is or how to input it on a calculator.
Conventional wisdom tells me that is "Function of X equals Sine of X times Line of X", but that would be f(x)=sin(x)*x which does not accurately graph the given answers.
Any assistance?(8 votes)
- ln stands for the natural logarithm.
It is the logrithm with base e.
e is an irrational number like pi approximately equal to 2.71828(18 votes)
- Is it possible to calculate such things without using a calculator within a reasonable time? Or is it almost impossible?(5 votes)
- If you don't use a calculator for calculating sin(x) or cos(x) you'll need a trig table -- that's what people used before calculators were invented. There's really no advantage to using a trig table that I can see. So you pretty much have to use a calculator for sin(x) and cos(x). The rest you might be able to do by hand in a reasonable amount of time, but there's really no reason to.(4 votes)
- How come when I'm in radian mode in a calculator,and when I need to input 1.0001 into 1/In(x)-1/x-1,how come I get something like 99985.124328663 instead of something like 0.9999932534?Whenever I try this,it always happens.IT'S DRIVING ME CRAZY!Can someone help me?(5 votes)
- How can you do these by hand(3 votes)
- You will learn how to do that as you advance in the course. There are mathematical ways to get the exact limit (if it exists) by direct means, not by estimations. But, you need to understand the concept of a limit before moving onto that material (and it eventually gets very advanced).(6 votes)
- How do you get a graphical calculator on your computer or iPads?(3 votes)
- Why is it important to know whether x is approaching to 0 from positive or not? then, if x is approaching 0 from negative is the x->0^ - underneath limit?(2 votes)
- Consider the equation
x/|x|. Everything to the right of x=0 will give you the answer 1, but everything to the left of x=0 will give you -1. So, the limit as x approaches zero is one if you approach from positive and -1 if you approach from negative, but if you don't specify which side you're approaching from, then the limit doesn't exist. It's important to specify which side you're approaching from if the limit on either side will be different.(4 votes)
- In the exercise after this video I came across the problem: f(t) = t^6-1/t^3+1. All had been well and good until this one, as the values in the table disagreed with what my calculator told me. For instance, in the table, when t= -1.01, f(t) = −2.030301, but my calculator told me otherwise, and so did just working it out by hand. (By hand I mean with the aid of a non-graphing calculator).
Here's a link to the calculator I used, for your convience with -1.001 already plugged in: https://www.desmos.com/calculator/quctv6lflq(3 votes)
- Your answer is correct. You may have had a problem with the calculator misinterpreting what you meant if you did not use parentheses.
Here's the computation:
= (t³-1), where t ≠ -1
plugging in t=-1.01
= (-1.01)³ - 1
= -1.030301 - 1
= -2.030301(4 votes)
- Unfortunately, the university I attend does not allow for the use of any calculators on exams in Calculus. Any advice on how to do homework? Should I or should I not use a calculator? The graphing calculator helps with visualization and long division and multiplication take a very long time to write out and such and converting decimals to fractions... huge numbers when you start dealing with ∆y/∆x(1 vote)
- You just need to make sure you know the material well enough to do the work efficiently without a calculator. You need to be able to easily spot singularities, asymptotes, vertices, end behavior, etc.
You should also know squares of all integers at the very least through 25. You should know the cubes at least through 10. And you should be able to do this whether you're allowed a calculator or not.
For example, when you see ∜1250 you should easily see that is 5∜2 because you know that 5⁴ = 625 and 625*2 = 1250.
I took this subject quite a long time ago, before even the best calculators would have been of much help anyway. It isn't easy material, of course, but I honestly think your university has a point in not allowing the calculator on the exams. The calculators have become so advanced that one of the better models could solve a problem you had no idea how to solve -- we don't need to have scientists and engineers who don't know what the math means and do the math their jobs require, but who just got through university by using a really good calculator.(7 votes)
- [Instructor] Consider the table with function values for f of x is equal to x squared over one minus cosine x at positive x-values near zero. Notice that there is one missing value in the table. This is the missing one right here. Use a calculator to evaluate f of x at x equals 0.1 and enter this number in the table rounded to the nearest thousandth. From the table, what does the one-sided limit, the limit as x approaches zero from the positive direction of x squared over one minus cosine of x appear to be? So let's see what they did. They evaluated when x equals one, f of x equals 2.175. When x gets even a little bit closer to zero, and once again, we're approaching zero from values larger than zero. That's what this little superscript positive tell us. We're at 0.5, and we're at 2.042. Then when we get even closer to zero, 0.2, f of x is 2.007. And so I'm guessing when I'm getting even closer, it's gonna be even closer to two right over here. But let's verify that. Get my calculator out. So I wanna evaluate x squared over one minus cosine of x when x is equal to 0.1. So the first thing I wanna actually, let me verify that I'm in radian mode 'cause otherwise I might get a strange answer. So I am in radian mode. And so let me evaluate it. So I'm going to have 0.1 squared over, let me do divide it by one minus cosine of 0.1. And this gets me 2.0016. And let's see, they want us to round to the nearest thousandth. So that'll be 2.002. Type that in. 2.002. And so it looks like the limit is approaching two. It's not approaching 2.005. We just crossed 2.005 from 2.007 to 2.002. So let's check our answer. And we got it right. And I always find it fun to visualize these things, and that's what a graphing calculator is good for. It can actually graph things. So let's graph this right over here. So go into graph mode. Let me redefine my function here. So let's see. It's going to be x squared divided by one minus cosine of x. And then let me make sure that the range of my graph is right, so I'm zoomed in at the right part that I care about. So let me go to the range. And let's see. I care about approaching zero from the positive direction, but as long as I see values around zero, I should be fine. But I could actually zoom in a little bit more, so I can make them my minimum x-value. I don't know, let me make it negative one. Let me make my maximum x-value. The maximum x-value here is one, but just to get some space here I'll make this 1.5. So the x-scale is one. Y-minimum, it seems like we're approaching two. So the y-max can be much smaller, let's see. We'll make y-max three. And now let's graph this thing. So let's see what it's doing. It looks like... And you see here whether, and actually it looks like whether you're approaching from the positive direction or from the negative direction, it looks like you're approaching, the value of the function approaches two. But this problem, we're only caring about as we have x-values that approaching zero from values larger than zero. So this is the limit. This is the one-sided limit that we care about with the two shows up right over here as well.