Calculus Limits

Can someone help me with Limits in AP Calculus AB?

An example problem-

Lim f(x)= (tan(x))/x
x -> 0

How can I make this into a "well behaved" function and solve for x to find the limit?

 

Do you have a graphing calculator?

 

Graphing calculators dont teach you anything and are illigal in exams.

Its been 2 years since I did all this shit, from what i remember is looks harder then it is. Try googling calculas limits or check out www.boredofstudies.com. yes i did mean bored, not board.

 

I tried googling but I can't find any website that would explain it to me well enough for me to totally grasp the concept.

I was hoping someone could help me so I could ask questions for anything I don't understand.

thequestionmark, you cannot solve this limit with a graphing calculator. It may aid you in achieving the end result, but it cannot solve it for you since the limit at 0 is undefined which means you cannot divide by zero. This limit can be solved by somehow rearranging the equation to another equation that is the same but isn't undefined at 0.

 

I haven't done this stuff yet, but I'll have a go at it:

(tanx)/x
=(sinx/cosx)/x
=(sinx/cosx)/x * (cosx/cosx)
=(sinxcosx/cosx)/(x*cosx)
=(sinx)/(x*cosx)
=(sinx)/x * 1/cosx

lim x->0 (sinx)/x = 1 (this is a handy identity)
(cos0) = 1

Since you no longer have a 0 as denominator, do your sum:

1 * 1 = 1.

I might be completely wrong, because as I say, I haven't done this stuff before. Hope it helps somehow

EDIT: I know my brackets are all in the wrong places, they should be around the x terms. But I think the maths itself is good.

 

http://www.physicsforums.com/archive/index.php/t-165896.html

There you go

 

http://www.clarku.edu/~djoyce/trig/identities.html - read the first identify off there.

Code:

 tan t = sin t /cos t

What do you think cosx/cosx simplifies to? One.

All the same, answer might be wrong. Someone smart please check this out?

 

oh shit my bad...

i read it cosx/sinx

and my fucking idiotic fat teacher said that Tanx = Cosx*Sinx

apologies ):

 

You can just use a t-84 calc for this...

 

I found the answer after a while of contemplating this problem and looking through my calculus book. (My teacher isn't too helpful.)

((sinx)/(cosx))/x = ((sinx)/x)(1/cosx) = (1)(1) = 1

Here is the correct solution.

Lim f(x)= (tan(x))/x
x -> 0

(tan(x))/x = ((sin(x))/(cos(x))/x
((sin(x))/(cos(x))/x = ((sin(x))/x)(1/cos(x))
((sin(x))/x)(1/cos(x)) = 1*1
1*1 = 1

The limit of the function as x approaches 0 is 1.

You guys were pretty much on the right track. Thanks for your help.

 

No problem and what grade are you in cause i havent learnt this yet

 

Not really... I'll repeat what I said to thequestionmark.

 

I'm in calculus 2.

Just because you don't know how to use your calculator, it doesn't mean that your calculator is incapable of solving something as simple as a limit... if your calculator can find values within .000001 of the limit, you can surely find what the limit is. As a matter of fact, if you DON'T know how to use this, you will find yourself in a far deeper hole in the AP. Just an FYI for the idiots who don't know this, there ARE sections of the AP which REQUIRE use of a graphing calculator, and most standardized approve of it (minus the ti89).

Duh.

To me more specific, since you obviously don't know. Input the equation into the Y= part of the calculator. You can either use tableset/table, or Vars, Y-Vars, Y1 input (limit-.00001).

 

so um?
SOHCAHTOA?

 

thequestionmark, I think you have misunderstood my problem.

You are wrong.

Lim f(x)= (tan(x))/x
x -> 0

Input y= (tan(x))/x and look at the value for 0
You will get not get a value for y. It will be undefined because you cannot divide by zero. In order to solve that equation, you have to rearrange it so you don't divide by zero. Like this:

As you can see, I changed tan(x) into (sin(x)/x)*(1/cos(x)).
Since you say are in Calculus 2 you should know (sin(x)/x)*(1/cos(x)) equals 1 and 1 respectively if you input zero for x. 1*1=1, limit of (tan(x))/x as x approaches 0 is 1.

Forgive my frankness, but if have indeed passed Calculus and you didn't understand this (obviously, since you said you CAN just use a calculator to solve this) then I guess you are in "a far deepet hole" for Calculus 2.

As I will say again, a calculator may AID you in achieving but WILL NOT always give you the answer alone.

 

You're an idiot. Flat out IDIOT.

Read it again moron. You can't find the value AT 0, but with a calculator, you can find the value of .0000001. This value, due to the approximation function of your calculator, will essentially EQUAL the limit. The only exception to these are when the limit approaches infinity (in which you will get a very high number), when the limit is different from either side (limit cannot exist), or in an oscillation equation (where the limit cannot exist).

Do I really need to spell this out to you? Are you really that freaking stupid?

If you need proof. DO EXACTLY as I say before you spew a load of bullshit out again.

1. Hit Y=
2. Y1= (tan(x))/x
3. Hit mode, then clear
3. Hit vars, switch to Y-vars, hit function, hit "Y1"
4. Enter Y1 (.00001)
5. Hit enter

End Result: 1

Typically, you must also check Y1 (-.99999) to make sure the number is the same as Y1 (.000001)

Easier method is to use table with .00001 intervals, shows limits from both sides.

Gf Idiot, I learned how to use my Calculator in Alegebra 2....
/endrant

 

Tell me, how do you know the limit is 1 in the first place? Even if you graph the function on your calculator you won't be able to see the "hole".
Your hindsight bias makes you know that the limit is at 1.

How do you know to put that in without knowing what the limit really is?
Either you just randomly guess what the limit is (which will take a while). Or you can solve for x in a "well-behaved" function. In order to solve such a function, the function cannot be undefined at 0 (cannot divide by 0).
Such as in the problem:
Lim f(x)= (tan(x))/x
x -> 0

Solving for x is much quicker than the former method.

The correct way to find the limit in a "well-behaved" function which is undefined at 0 is to rearrange it so it isn't undefined at 0 anymore (not dividing by 0).

My method is correct.

You need to correct your own temper, how can such a simple dispute bring out your vulgar side?

 

Right on

 

You clearly don't understand how limits work. The value CANNOT be solved at that point, but CAN be solved within .00001 of the value. There is NO hindsight bias, you know to enter those values because finding it at 0 is a GIVEN.

Hindsight bias would be entering in 0 + 1 and getting one. I picked values .000001 and - .99999 becaues of their closeness to 0, NOT because they solve for one.

You need to correct your incompetence.

Viper, on the contrary. Understanding Calculus is the precursor to using your calculator. Real life applications all involve using things "unsolvable" by hand.

 

Possibly I did calculus for A-Level, add my msn.