Limits (one algebreic, two Trig)

[Alethia]

Well I just started to learn limits in my calculus class, and I'm getting it fairly well, but I'm a bit confused on these three:

1) the limit of (x^5 - 32)/(x-2) as x approaches 2.
The problem I have here is factoring the (x^5 - 32) part. Once I can get that factored I think I can handle the rest. :D

2) what is the limit of xsecx as x appraoches pi.
I simplified that to x(1/cosx) but that really got me nowhere.

3) and, last but not least, what is the limit of (1-tanx)/(sinx-cosx) as x approaches pi/4?
I broke this one down to [(2 sin 2x)/2x)(mx/3 sin 3x)], but I don't know where to go from there.

The two basic trig fxns I learned are:
the limit of sin x/x as x approaches 0 is equal to one
and
the limit of (1-cos x)/x as x approaches 0 is equal to zero

If anybody coiuld please lend me their wisdom, lead me into the right direction, I would be forever grateful. Thank you!

 

[Pyrrhus]

1) the limit of (x^5 - 32)/9x-2) as x approaches 2.
The problem I have here is factoring the (x^5 - 32) part. Once I can get that factored I think I can handle the rest. :D

Did you try substituting for 2 before trying to transform the expression, you must first try and see if it will create an inderterminate form, i see 0/16 which is 0

 

[Alethia]

Did you try substituting for 2 before trying to transform the expression, you must first try and see if it will create an inderterminate form, i see 0/16 which is 0

YEah, I typed the problem wrong, but I edited it. And yes I did plug in, and got 0/0 so now that means I need to factor further, so... heh.

 

[Tide]

You might recognize (y^n - q)/(y-1) as the sum of a geometric series and go from there. If not, try dividing x - 2 directly into x^5 - 32. You should see a pattern emerge even before you finish it! :-)

 

[recon]

Have you not learned the http://www.math.hmc.edu/calculus/tutorials/prodrule/ and http://www.math.hmc.edu/calculus/tutorials/quotient_rule/? If you know these two rules, solving the problems you posted will only be a matter of plugging in the value of x.

 

[HallsofIvy]

Have you not learned the http://www.math.hmc.edu/calculus/tutorials/prodrule/ and http://www.math.hmc.edu/calculus/tutorials/quotient_rule/? If you know these two rules, solving the problems you posted will only be a matter of plugging in the value of x.

No, it's not. Problem 2 can be handled that way, but problems 1 and 3 both have denominators that are 0 at the "target" point and so the quotient rule does not apply (in fact, it doesn't apply in all the interesting cases!).

Alethia was correct- you need to factor (x⁵- 32).

In general, xⁿ-aⁿ can be factored as
(x-a)(x^n-1+ ax^n-2+ a²x^n-3+...+a^n-2x+ a^n-1).
x⁵- 32= (x-2)(x⁴+ 2x³+ 4x²+ 8x+ 16).

I'm not at all sure how Alethia got the "3x" and "2x" in the 3rd problem (nor what "m" is!). Since tan(x)= sin(x)/cos(x), 1- tan(x)= 1- sin(x)/cos(x)=
(cos(x)- sin(x))/cos(x) so (for x not &pi/4)

(1- tan(x))/(sin(x)- cos(x))= ((cos(x)- sin(x))/cos(x))(1/(sin(x)- cos(x)) and that should be easy.

 

[Alethia]

Thank you everyone! I think I just factored wrong on the first one. I used synthetic, but I kept getting the wrong answer, but I get it now. :D

 

[recon]

Woops, sorry for the blunder I made. I'll try to be more careful in the future.