Question on calc limits

[jnmnh]

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I have a question in my study guide and I can't figure out how the answer was reached.

Problem:
lim x -> 1 (x^3 - x^2) / (1-x^3)

Answer:
-1/3

Is there a math master out there that can explain this to me? I feel like I am just missing how to factor it.

Are limits on the PCAT only inifinite limits and 0/0 limits?

And finally while I have help what is d(ln 2x)/dx?
Thanks!

 

[yankies90]

Ok I got it. I will help with the limit problem. Numerator will give you 0 when you plug in 1 righ? while denominator gives 0 too. What you have to do at this situation is take derivative of both numerator and denominator and plug in the number which is 1 again.

It becomes 3x^2 - 2x / -3x^2 when I took derivative of both numerator and denominator separately. and just simply plug in 1 into both top and botton and will give you the answer 🙂

and I think d(ln 2x)/dx gives you 1/x if I remember my stuffs correctly.

 

[justhanging]

Ok I got it. I will help with the limit problem. Numerator will give you 0 when you plug in 1 righ? while denominator gives 0 too. What you have to do at this situation is take derivative of both numerator and denominator and plug in the number which is 1 again.

It becomes 3x^2 - 2x / -3x^2 when I took derivative of both numerator and denominator separately. and just simply plug in 1 into both top and botton and will give you the answer 🙂

and I think d(ln 2x)/dx gives you 1/x if I remember my stuffs correctly.

The derivative is 1/2x *2

use the chain rule.

 

[delano2000]

The derivative is 1/2x *2

use the chain rule.

Answer would still come out to be 1/x when you simplify.

 

[FormerRT]

Answer would still come out to be 1/x when you simplify.

when did you take calc? did you do it at SPC?

 

[justhanging]

Answer would still come out to be 1/x when you simplify.

True but then you just learn the derivative of ln F(x) is just one over x, which is not the case.

 

[delano2000]

when did you take calc? did you do it at SPC?

I took it this past semester and yes at SPC.

 

[delano2000]

True but then you just learn the derivative of ln F(x) is just one over x, which is not the case.

Lol I get that but I was looking at it in terms of simplifying. I know you would also have to take the deriv. of the x term as well hence chain rule. 👍

 

[NewBee12]

Hi
you dont need to use derivatives here, all you have to do is factor:

(x^3-x^2)/(1-x^3) = x^2(x-1)/(1-x)(1+x+x^2)

[ a^3-b^3 =(a-b)(a^2+ab+b^2) ]
then cancel (x-1) and (1-x) you will be left with -1
now you have
lim x--->1 -x^2/(1+x+x^2) plug in 1 and you will get -1/3

good luck

 

[delano2000]

Hi
you dont need to use derivatives here, all you have to do is factor:

(x^3-x^2)/(1-x^3) = x^2(x-1)/(1-x)(1+x+x^2)

[ a^3-b^3 =(a-b)(a^2+ab+b^2) ]
then cancel (x-1) and (1-x) you will be left with -1
now you have
lim x--->1 -x^2/(1+x+x^2) plug in 1 and you will get -1/3

good luck

Isn't finding the derivative and pluging in the limit so much easier? Whichever way is faster for you OP.