# Compute the Limit: Question 2

• Sep 20th 2011, 07:46 PM
sparky
Compute the Limit: Question 2
I'm stuck on this one:

$\text{Compute } \lim_{u \rightarrow 1 } \frac{u^4-1}{u^3-1}$

$= \frac{1^4-1}{1^3-1}$

$= \frac{0}{0 } \text{Indeterminate}$

$\lim_{u \rightarrow 1 } \frac{u^4-1}{u^3-1}$

$= \lim_{u \rightarrow 1 } \frac{(u-1)(u+1)(u-1)(u-1)}{(u-1)(u+1)(u+1)}$

$= \lim_{u \rightarrow 1 } \frac{(u-1)(u-1)}{(u+1)}$

$= \frac{(1-1)(1-1)}{(1+1) }$

$= \frac{0}{2}$
• Sep 20th 2011, 08:12 PM
TKHunny
Re: Compute the Limit: Question 2
Not good, Sparky. Please factor numerator and denominator again. Both are incorrect.

Not good, Sparky. Why did you substitute the limit value into a function you do NOT know is continuous? Never do that.

Do a little less manipulating little symbols and a lot more thinking.
• Sep 24th 2011, 05:14 PM
sparky
Re: Compute the Limit: Question 2
Ok, here is another attempt at the following question:

$\text{Compute } \lim_{u \rightarrow 1 } \frac{u^4-1}{u^3-1}$

$u^4-1 = (u^2+1)(u^2-1)$

$u^3-1 = (u-1)(u^2+u+1)$

$\lim_{u \rightarrow 1 } \frac{(u^2+1)(u^2-1)}{(u-1)(u^2+u+1)}$

Where did I go wrong with my factorization?
• Sep 24th 2011, 05:38 PM
Deveno
Re: Compute the Limit: Question 2
so far, so good. you won't be able to factor $u^2+1$ further, but you CAN factor $u^2-1$. try that, and see if something cancels...
• Sep 24th 2011, 06:02 PM
sparky
Re: Compute the Limit: Question 2
Quote:

Originally Posted by Deveno
so far, so good. you won't be able to factor $u^2+1$ further, but you CAN factor $u^2-1$. try that, and see if something cancels...

$\text{Compute } \lim_{u \rightarrow 1 } \frac{u^4-1}{u^3-1}$

$u^4-1 = (u^2+1)(u^2-1)= (u^2+1)(u+1)(u-1)$

$u^3-1 = (u-1)(u^2+u+1)$

$= \lim_{u \rightarrow 1 } \frac{(u^2+1)(u+1)(u-1)}{(u-1)(u^2+u+1)}$

$= \lim_{u \rightarrow 1 } \frac{(u^2+1)(u+1)}{(u^2+u+1)}$

$= \frac{1+1+1+1}{1+1+1}$

$= \frac{4}{3}$

Is this correct?
• Sep 24th 2011, 06:18 PM
TKHunny
Re: Compute the Limit: Question 2
It is a beautiful thing.
• Sep 25th 2011, 06:29 AM
Deveno
Re: Compute the Limit: Question 2
Quote:

Originally Posted by sparky
$\text{Compute } \lim_{u \rightarrow 1 } \frac{u^4-1}{u^3-1}$

$u^4-1 = (u^2+1)(u^2-1)= (u^2+1)(u+1)(u-1)$

$u^3-1 = (u-1)(u^2+u+1)$

$= \lim_{u \rightarrow 1 } \frac{(u^2+1)(u+1)(u-1)}{(u-1)(u^2+u+1)}$

$= \lim_{u \rightarrow 1 } \frac{(u^2+1)(u+1)}{(u^2+u+1)}$

$= \frac{1+1+1+1}{1+1+1}$

$= \frac{4}{3}$

Is this correct?

your answer is correct, but the work is not. it is lucky for you that (1+1)(1+1) = 1+1+1+1.
• Sep 25th 2011, 04:01 PM
sparky
Re: Compute the Limit: Question 2
Quote:

Originally Posted by Deveno
your answer is correct, but the work is not. it is lucky for you that (1+1)(1+1) = 1+1+1+1.

My work is not correct? Where did I go wrong?
• Sep 25th 2011, 10:04 PM
Siron
Re: Compute the Limit: Question 2
In this step you're substituting wrong, you wrote:
$\lim_{u\to 1} \frac{(u^2+1)(u+1)}{u^2+u+1}=\frac{1+1+1+1}{1+1+1} =\frac{4}{3}$
It has to be:
$\lim_{u\to 1} \frac{(u^2+1)(u+1)}{u^2+u+1}=\frac{(1+1)(1+1)}{1+1 +1}=\frac{4}{3}$

Do you notice your mistake? Offcourse like Deveno said it's lucky that (1+1)(1+1)=1+1+1+1, but in other cases ...