Question

Find the limit for number 4 numerically. Show work.

Answer 1

The limit of just "4"?

Answer 2

not sure

Answer 3

Well, there has to be more to it than that. The limit of 4 is 4. The limit of 6 is 6, the limit of 8 is 8. Unless there's some sort of function or something, then I'm not sure what there is to add, unfortunately.

Answer 4

what's the limit of \[\frac{ (\sqrt{x+50)}-3 }{ x-4}\]

Answer 5

Okay, that makes more sense. As x approaches 4?

Answer 6

Rationalize it.

Answer 7

0

Answer 8

Well if x is approaching 0, then just plug in 0 for x and see what you get.

Answer 9

and if it's approaching 4?

Answer 10

Well, it looked like your problem said numerically. Numerically is different than algebraically. Numerically means you have to approximate by calculating the answer with x = 3.99 and x = 4.01 or something. Algebraically means we have to rationalize the numerator and actually try to manipulate it. Do you know which method you need to do?

Answer 11

not really

Answer 12

Well, then we should probably do it the algebraic way. Do you know how to rationalize the numerator?

Answer 13

If I were to rationalize the numerator, would the outcome be 0?

Answer 14

Nope.

Answer 15

You'd get an actual answer :3

Answer 16

Do I replace x with 4?

Answer 17

After you rationalize the numerator, yes.

Answer 18

Ok so I probably don't know how to rationalize?

Answer 19

Well, to rationalize the numerator you need to multiply the top and bottomof your fraction by the conjugate of whatever you are rationalizing. So we want to rationalize
\[\sqrt{x+50}-3\]The conjugate is just the same thing but with the middle sign changed. So this means we have to multiply top and bottom but
\[\sqrt{x+50}+3\]This means we need to be able to simplify this:
\[\frac{ \sqrt{x+50}-3 }{ x-4 }*\frac{ \sqrt{x+50}+3 }{ \sqrt{x+50}+3 }\]

Answer 20

Assuming I remembered the problem correctly xD

Answer 21

haha well actually it 5 and not 50. Typo :3

Answer 22

But same concept right?

Answer 23

Yes, same concept. Yeah, I remember you having it typed as 50 xD But yeah, same thing :3

Answer 24

Ok, time for simplifying haha

Answer 25

Lets see what ya get then :3

Answer 26

Well, I don't think that's possible...

Answer 27

Yeah, it still comes out to be 0/0

Answer 28

It wasnt 0/0 before because I thought it was 50 xD

Answer 29

So hold on, I was right earlier or ???

Answer 30

Kinda confused here haha

Answer 31

Well, this is a calc 1 class, correct?

Answer 32

yeah, sadly

Answer 33

Alright. The reason I ask is because there is a calculus 2 method that solves this problem, lol. So yes, we have to do this analytically since its calc 1. This means we have to choose numbers close to 4 and calculate them. This then lets us take an educated guess at what the limit is. So we would have to see what happens when x = 3.99 and what happens when x = 4.01 and then see if we can figure out the limit from there.

Answer 34

with that table? haha

Answer 35

Yeah, pretty much, lol.

Answer 36

well, that's the easy thing haha. And figure out if the limit from the left side and the limit from the right side approach it

Answer 37

Calc 2 is harder than Calc 3!!! That's crazy

Answer 38

Im only now starting calc 3 xD Calc 2 was easy for me O.o There was only one chapter that was kinda hard, I had np in it to be honest. But then again, im on here as a math tutor, not sure I count, lol.

Answer 39

My brother took up til Calc 3. He said Calc 2 is harder than Calc 3. You fly by in it. Sadly, I wasn't blessed with math skill... Oblviously.

Answer 40

The calc 2 method involves something called l'hopitals rule. It involves separately takign the derivative (something youll learn very soon) of the numerator and dividing it by the derivative of the denominator.
For this problem the numerator:
\[\frac{ d }{ dx }\sqrt{x+5}-3=\frac{ 1 }{ 2\sqrt{x+5} }\]
Divide by the derivative of the denominator, which is only 1 actually. So that means we are only left with the above, in which we just plug in 4 for x : )
And I hope thats good for me xD If i had an easy time in calc 2 then calc 3 better be easy >.<

Answer 41

Wow, that's a lot to know. I've heard of derivatives. I learned it slightly but I totally forgot it. So that's the l'hopitals rule, in Calc 2 and not 1 right?

Answer 42

Right. It basically says that the limit of a function is the same as the limit of the derivative of the numerator divided by the derivative of the denominator. There is one catch, though. It can only be used if plugging in the number for x results in 0/0 or infinity/infinity or one of the "indeterminant" forms. This problem just happenes to be a 0/0, so it works in this case. And yeah, calc 2. But if you can do derivatives well, then its an easy concept.

Answer 43

Then it's actually pretty simple. Let's just hope I find derivatives easy. Something like f(x)=2x^2-16x+35

Answer 44

The derivative of that would be something like f(x)= 4x-16

Answer 45

They can be pretty tricky, but you start to figure out how to do them and what the actual process is and then ts just a matter of paying attention. And yeah, it would xD

Answer 46

I only remember that problem and how to do it haha

Answer 47

Yeah xD Now this is a bit of a ridiculous example, but if you learn derivatives well enough, this would be something that I guess you could say is a "challenge" problem for someone who has finished calc 1. A lot of those people would probably get it wrong, but it's something that supposed to be known how to do xD
\[\frac{ x ^{2}\sqrt[4]{x ^{3}-xe ^{x}} }{ \sqrt{\sqrt{x ^{2}-6}+\sqrt{lnx}} }\]

Answer 48

That does look like a challenge haha

Answer 49

Well, basically its a test of being careful and a test of if you truly understand how to doderivatives. If you know the rules and the process well enough, its doable. Its an annoying problem, but doable. Basically, its one of those problems that either you can doit or you cant. You either get all the rules of derivatives or ya dont xD

Answer 50

I see. So it's fairly simple IF you know the rules. If not then practically impossible?

Answer 51

Impossible if you dont, lol. If you do know the rules, its still a very ridiculous problem to do. i got the answer and its insane, haha. But like I said, either you know how to do it or you dont.

Answer 52

Math can be too much at some point. Can you give me a real world problem where you would have to find the derivative of something? haha

Answer 53

Well, I can make it real, but I cant come up with the greatest example that me and you would be using. But like if you know the dimensions of like a balloon. And you're pumping air into the balloon. If you know the rate youre pumping air in and you can maybe measure and say the balloon was a diamter of 10 inches after 10 seconds and 15 inches after 25 seconds, you can then calculate things like how fast is the radius changing, how fast is the volume changing, how fast is the surface area changing. if you know the balloon can only hold so much air you can calculate the amount of time itll take before it reaches that threshold.
Derivatives of measures of change. They find the rates of change of various things, it can be graphs or it can be real world things.

Answer 54

Nice work. That's actually a good example for it though. I get where you're going with it so that's good.

Answer 55

Mhm. A lot of the real fancy things that all the engineers and physicists and all that are doing usually use a mix of derivatives and integration. Integration is basically the reverse of differentiation. But a lot of the awesome stuff people measure and figure out use a mix of both in the same problem. But there are several things and only use one or only use the other. But yeah, Im gonna head out now, need to get some sleep, lol. Night :3

Answer 56

What a job they have. A lot of work then. Haha alrighty. Night :3