lim(x --> 0) (x^4 - 1) / (x^3 - 1) (tried these methods, but

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green_tea

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Hi guys, i need help agian :oops:

I have to solve this:

lim (x^4-1)/(x^3-1)
x-->1

I've been trying a lot of things, for example this:

(x^4-1)/(x^3-1) = (x - 1/x^3) / (1- 1/x^3)

or this:

(x^4-1)/(x^3-1) = (x-1)(x+1)(x^2+1)/(x^3-1)

I also tried to put x=1-h and then see what happends with the equation when h-->0 but everytime I get stuck with the dividing with 0 thing.
I heard that there's some method one can use with differentiation but we're supposed to solve this without that.
Does anyone know how to solve this? (You don't have to solve the whole thing for me, it's enough if you just tell me how to start...)
 
Re: lim(x --> 0) (x^4-1)/(x^3-1)

lim x-->0 [x^4-1]/[x^3-1] improper fraction, divide
lim x-->0 {x +[x-1]/[x^3-1] } factor denominator
lim x-->0 {x +[x-1] /[[x-1][x^2+x+1]]}
lim x-->0 {x+1/[x^2+x+1]
lim x-->0[0 + 1/[1]
lim x-->0 of[x^4-1]/[x^3-1]= 1 answer
Arthur
 
Re: lim(x --> 0) (x^4-1)/(x^3-1)

green_tea said:
Hi guys, i need help agian :oops:

I have to solve this:

lim (x^4-1)/(x^3-1)
x-->1

=lim (x^2+1)(x-1)(x+1)/[(x-1)(x^2 + x + 1)]
x-->1

=lim (x^2+1)(x+1)/[(x^2 + x + 1)]
x-->1

Now take the limit by replacing x with 1



I've been trying a lot of things, for example this:

(x^4-1)/(x^3-1) = (x - 1/x^3) / (1- 1/x^3)

or this:

(x^4-1)/(x^3-1) = (x-1)(x+1)(x^2+1)/(x^3-1)

I also tried to put x=1-h and then see what happends with the equation when h-->0 but everytime I get stuck with the dividing with 0 thing.
I heard that there's some method one can use with differentiation but we're supposed to solve this without that.
Does anyone know how to solve this? (You don't have to solve the whole thing for me, it's enough if you just tell me how to start...)
Click to expand...
 
Re: lim(x --> 0) (x^4-1)/(x^3-1)

Hello, green_tea!

\(\displaystyle \lim_{x\to1}\frac{x^4-1}{x^3-1}\)
Click to expand...

Did you try factoring and reducing?


\(\displaystyle \frac{x^4-1}{x^3-1} \;=\;\frac{(x^2-1)(x^2+1)}{(x-1)(x^2+x+1)} \;=\;\frac{(\rlap{/////}x-1)(x+1)(x^2+1)}{(\rlap{/////}x-1)(x^2+x+1)} \;=\;\frac{(x+1)(x^2+1)}{x^2+x+1}\)


\(\displaystyle \text{Therefore: }\;\lim_{x\to1}\frac{(x+1)(x^2+1)}{x^2+x+1} \;=\;\frac{2\cdot2}{1+1+1} \;=\;\frac{4}{3}\)

 
Re: lim(x --> 0) (x^4-1)/(x^3-1)

the problem was limit as x approachs 0, not 1

lim x-->0 [x^4-1]/[x^3-1] improper fraction , divide
lim x--> [x+ [x-1] / [x^3-1] ] factor denominator
lim x-->0 [x^4-1]/[x^3-1] = lim x-->0 {x+ [x-1]/[ x-1][x^2+x+1]
lim x-->0 [x^4-1]/[x^3-1]= 0+1/[0+0+1]
lim x-->0 [x^4-1] / [x^3-1] = 1 answer
Arthur
 
Re: lim(x --> 1) (x^4-1)/(x^3-1)

Thanx all for responding so fast!!! I get it now :)
(Arthur, it sholud be x ---> 1 , I noticed now that I wrote 0 in the subject, but it sholud be lim x---> 1 sorry..:/ )
 
Re: lim(x --> 0) (x^4-1)/(x^3-1)

arthur ohlsten said:
the problem was limit as x approachs 0, not 1
Arthur
Click to expand...
Arthur, lot of times students write one thing in the subject line - but "change-their-mind" in actual post (2 nd. mistake this year - Dennis is keeping records - his new job after crossing 5000 post mark).

However, for limit x --> 0, wouldn't just evaluating the function at x = 0 be sufficient?
 
Re: lim(x --> 0) (x^4-1)/(x^3-1)

The math becomes the same for lim x-->1 as for lim x-->0 but answer is 1+1/3

lim x-->1[x^4-1] /[x^3-1] improper fraction divide
lim x-->1 [x^4-1] / [x^3-1] = lim x-->1 { x + [x-1]/[x^3-1] } factor denominator
lim x-->1 [x^4-1] / [x^3-1] = x +( [x-1] / { [x-1][x^2+x+1] } )
lim x-->1 [x^4-1]/[x^3-1] = x+ 1/[x^2+x+1]
lim x-->1 [x^4-1] /[x^3-1]=1+1/3 answer
Arthur
 
Re: lim(x --> 0) (x^4-1)/(x^3-1)


Hello, Arthur!


If it really was \(\displaystyle x\to0\), the problem is trivial . . . no algebra is required.

\(\displaystyle \lim_{x\to0}\frac{x^4-1}{x^3-1} \;=\;\frac{0^4-1}{0^3-1} \;=\;\frac{-1}{-1} \;=\;1\)

 
Re: lim(x --> 0) (x^4-1)/(x^3-1)

I agree the problem is trivial.
I don't know what the instructor was attempting to teach. The approach I showed the student would be correct regardless if the limit was 0,or 1.

He mentioned he wasn't to do it by L'Hospitals rule.

I first showed him that he had a improper fraction and to reduce the function to a proper fraction. This also was not neccessary but a good habit for students to know

I then showed him the Basic Algorithim of Arithmetic, " any number can be written as a product of primes". This should help the student in factoring.

I agree none of this is neccessary, but what was the instructor trying to teach? I wasn't sure
Arthur
 
Re: lim(x --> 0) (x^4-1)/(x^3-1)

x --> 0 must be typo in the title ... the original post has x approaching 1 in the text part of the problem.
 
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