need help with an easy limit

**newby** · December 17th 2009, 02:14 AM

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

I tried with L'Hospital's rule but it seems it doesn't work (it still gives me 0/0... Maybe I didn't apply it well?). Any suggestions?

Thanks in advance

!

[Sorry if it's a dumb question, but I am not very good at math...

]

**Raoh** · December 17th 2009, 02:37 AM

Originally Posted by newby

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

I tried with L'Hospital's rule but it seems it doesn't work (it still gives me 0/0... Maybe I didn't apply it well?). Any suggestions?

Thanks in advance

!

[Sorry if it's a dumb question, but I am not very good at math...

]

$\lim_{x\to 1}\frac{x^3-3x +2}{x^4-4x + 3}$ = $\lim_{x \rightarrow 1 } \frac{ 3x^2-3}{4x^3-4 }$ = $\lim_{ x \rightarrow 1} \frac{3 }{4 } \left ( \frac{ x^2-1}{x^3-1 } \right )$ = $\lim_{ x \rightarrow 1} \frac{ 3}{ 4} \left ( \frac{ x+1}{x^2+x+1 } \right )$ = $\frac{ 3}{4 } \left ( \frac{2 }{3 } \right ) = \frac{ 1}{ 2}$
hope i'm right

**newby** · December 17th 2009, 02:40 AM

Originally Posted by Raoh

$\lim_{x\to 1}\frac{x^3-3x +2}{x^4-4x + 3}$ = $\lim_{x \rightarrow 1 } \frac{ 3x^2-3}{4x^3-4 }$ = $\lim_{ x \rightarrow 1} \frac{3 }{4 } \left ( \frac{ x^2-1}{x^3-1 } \right )$ = $\lim_{ x \rightarrow 1} \frac{ 3}{ 4} \left ( \frac{ x+1}{x^2+x+1 } \right )$ = $\frac{ 3}{4 } \left ( \frac{2 }{3 } \right ) = \frac{ 1}{ 2}$
hope i'm right

Thanks a lot! I feel more clever now

!
Thanks again

!

**earboth** · December 17th 2009, 03:05 AM

Originally Posted by newby

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

I tried with L'Hospital's rule but it seems it doesn't work (it still gives me 0/0... Maybe I didn't apply it well?). Any suggestions?

Thanks in advance

!

[Sorry if it's a dumb question, but I am not very good at math...

]

Here comes a slightly different way:

As long as the quotient is $\frac00$ you can apply de l'Hopital's rule:

$\lim_{x\to 1}\frac{x^3-3x +2}{x^4-4x + 3} = \lim_{x\to 1}\frac{3x^2-3}{4x^3-4} = \lim_{x\to 1}\frac{6x}{12x^2} = \frac6{12} = \frac12$

**newby** · December 17th 2009, 03:26 AM

Originally Posted by earboth

Here comes a slightly different way:

As long as the quotient is $\frac00$ you can apply de l'Hopital's rule:

$\lim_{x\to 1}\frac{x^3-3x +2}{x^4-4x + 3} = \lim_{x\to 1}\frac{3x^2-3}{4x^3-4} = \lim_{x\to 1}\frac{6x}{12x^2} = \frac6{12} = \frac12$

Thank you too! I didn't know I could use the rule many times!

**Raoh** · December 17th 2009, 03:40 AM

Originally Posted by newby

Thank you too! I didn't know I could use the rule many times!

neither i

.

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