Compute the Limit: Question 2

I'm stuck on this one:

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

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

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

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

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

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

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

$\displaystyle = \frac{0}{2}$

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.

Re: Compute the Limit: Question 2

Ok, here is another attempt at the following question:

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

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

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

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

Where did I go wrong with my factorization?

Re: Compute the Limit: Question 2

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

Re: Compute the Limit: Question 2

Quote:

Originally Posted by

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

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

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

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

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

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

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

$\displaystyle = \frac{4}{3}$

Is this correct?

Re: Compute the Limit: Question 2

Re: Compute the Limit: Question 2

Quote:

Originally Posted by

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

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

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

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

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

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

$\displaystyle = \frac{4}{3}$

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.

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?

Re: Compute the Limit: Question 2

In this step you're substituting wrong, you wrote:

$\displaystyle \lim_{u\to 1} \frac{(u^2+1)(u+1)}{u^2+u+1}=\frac{1+1+1+1}{1+1+1} =\frac{4}{3}$

It has to be:

$\displaystyle \lim_{u\to 1} \frac{(u^2+1)(u+1)}{u^2+u+1}=\frac{(1+1)(1+1)}{1+1 +1}=\frac{4}{3}$

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 ...