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Question:

Find the limit: $$\lim_{x \to 1}\frac{x^4 - 3x^3 + 2}{x^3 -5x^2+3x+1}$$

The denominator can be simplified to: $$(x-1)(x^2+x)$$ However, I am unable to factor the numerator in a proper manner (so that $(x-1)$ will cancel out)

I know upon graphing that the limit is $5\over4$. What should I do here?

Note: To be done without the use of L'Hospital Rule

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  • $\begingroup$ Do you know L'Hospital? $\endgroup$ – Vim Feb 24 '15 at 12:13
  • $\begingroup$ Did you try the Euclidean division? $\endgroup$ – user63181 Feb 24 '15 at 12:13
  • $\begingroup$ @Vim Forgot to mention, but we are supposed to do this without the L'Hospital rule. $\endgroup$ – Gummy bears Feb 24 '15 at 12:14
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    $\begingroup$ Your simplication of denominator is incorrect. $x=0$ is not a root of it. $\endgroup$ – drhab Feb 24 '15 at 12:16
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    $\begingroup$ Yes. $x=1$ is a root of numerator and denominator. $\endgroup$ – drhab Feb 24 '15 at 12:18
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Here are the steps $$\lim_{x \to 1}\frac{x^4 - 3x^3 + 2}{x^3 -5x^2+3x+1}$$ $$= \lim_{x \to 1}\frac{(x-1)(x^3- 2x^2-2x-2)}{(x-1)(x^2-4x-1)} $$ $$= \lim_{x \to 1}\frac{x^3- 2x^2-2x-2}{x^2-4x-1} $$ $$= \frac{1- 2-2-2}{1-4-1} = \frac{1- 6}{1-5} = \frac{5}{4} $$

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Using the Euclidean division we get

$$\begin{array}\\x^4-3x^3+2&\Bigg|&x-1\\ -(x^4-x^3)&\Bigg|&x^3\\ =-2x^3+2&\Bigg|&-2x^2\\ -(-2x^3+2x^2)\\ =-2x^2+2&\Bigg|&-2x\\-(-2x^2+2x)\\=-2x+2&\Bigg|&-2 \end{array}$$

so we find that $$x^4-3x^3+2=(x-1)(x^3-2x^2-2x-2)$$

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You may use Horner here. Since $x=1$ is a root of the nominator (you can check that easily) then you can deduce that down by applying a long division.

Hence your limit is deduced down to: $$\lim_{x\rightarrow 1}\frac{x^4-3x^3+2}{x^3-5x^2+3x+1}=\lim_{x\rightarrow 1}\frac{(x-1)\left ( x^3-2x^2-2x-2 \right )}{\left ( x-1 \right )\left ( x^2-4x-1 \right )}=\frac{1-2-2-2}{1-4-1}=5/4$$

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  • $\begingroup$ @Tolaso LHR tells me it shoud be $\frac54$ why? $\endgroup$ – Vim Feb 24 '15 at 12:27
  • $\begingroup$ This is the best solution for someone who does not know many mathematics or for someone who is not very good handling mathematical quantities. $\endgroup$ – Konstantinos Gaitanas Feb 24 '15 at 16:52
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Hint: The numerator can be factorized as $$x^4-3x^3+2=x^4-1-3(x^3-1)=(x-1)((x^2+1)(x+1)-3(x^2+x+1))$$ and the denominator as $$x^4-5x^2+3x+1\\=x^4-1-(5x^2-3x-2)=(x-1)((x^2+1)(x+1)-(5x+2))$$

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