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I tried to factorize it but it doesn't work with it. Is there another way than factorising?

Thanks

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closed as off-topic by Henrik, Shailesh, Leucippus, Pragabhava, JMP Oct 15 '16 at 4:43

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  • $\begingroup$ is your numerator correct $\endgroup$ – hamam_Abdallah Oct 14 '16 at 20:38
  • $\begingroup$ sorry I didn't notice that there is an extra x. $\endgroup$ – John Oct 14 '16 at 20:40
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    $\begingroup$ use Hospital rule $\endgroup$ – hamam_Abdallah Oct 14 '16 at 20:41
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Factorizing does work. We have $$\frac{x^3 - 7x + 6}{3x^2 - 8x + 4} = \frac{(x - 2)(x^2 + 2x - 3)}{(x - 2)(3x - 2)}$$ so the limit is $$\frac{2^2 + 2\cdot 2 - 3}{3 \cdot 2 - 2} = \frac{5}{4}.$$

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  • $\begingroup$ finally an answer that makes sense, Thank. But how did you know that you have to use x-2 as a common factor? $\endgroup$ – John Oct 14 '16 at 20:50
  • $\begingroup$ Because you are taking the limit as $x \to 2$. But $2$ is a zero of the denominator, so you can't simply plug in $2$. Getting rid of the factor of $(x - 2)$ fixes the problem. $\endgroup$ – Ethan Alwaise Oct 14 '16 at 20:52
  • $\begingroup$ @John for any non-zero polynomial $P(x)$, $P(a) = 0 \iff (x-a) \text{ is a factor}$. This follows from the division theorem for polynomials. $\endgroup$ – GFauxPas Oct 14 '16 at 20:53
  • $\begingroup$ My bad, I studied polynomials before but it seems that I didn't pay attention to this theorem. I will revise it now. $\endgroup$ – John Oct 14 '16 at 21:02
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You may use L'Hôpital's rule.

Both numerator and denominator vanish for $x=2$, so you may give a try to the derivatives:

$$\lim_{x\to2}\frac{3x^2-7}{6x-8}=\frac{5}{4}$$

The previous limit exists and is finite, hence

$$\lim_{x\to2}\frac{x^2-7x+6}{3x^2-8x+4}=\frac{5}{4}$$

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$3x^2-8x+4 = (3x-2)(x-2), x^3-7x^2+6x = x(x-1)(x-6)$. You can see that the limit does not exist because the left and righthand limits at $x = 2$ are not the same, and one is $\infty$ while the other is $-\infty$.

Note: Due to your edit, I solved a different question, but it is still relevant and provides you with added experience on how to deal with non-existing limit.

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  • $\begingroup$ I am sorry it was a stupid mistake, but I totally appreciate your answer. $\endgroup$ – John Oct 14 '16 at 20:51
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let us factor.

for numerator:

$x^3-7x+6=$

$x^3-4x-3x+6=$

$x(x^2-4)-3(x-2)=$

$x(x-2)(x+2)-3(x-2)=$

$(x-2)(x(x+2)-3)$.

for denominator;

$3x^2-8x+4=$

$3x^2-6x-2x+4=$

$3x(x-2)-2(x-2)=$

$(x-2)(3x-2)$.

after simplifying, we get the limit

$\frac{5}{4}$.

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Since both the numerator and the denominator vanish at $2$, factorization must work, because both terms are divisible by $x-2$.

In Italy it's customary to use the “Ruffini scheme”, write the coefficients of the polynomial, with $0$ where a term is missing; separate the constant term; then push the first coefficient at the bottom, multiply it by $2$ and write the result below the next coefficient, sum with it and write the result on the bottom line, then repeat. In the last column we get the remainder, that should be $0$ when $x-2$ is a factor. $$ \begin{array}{r|rrr|r} & 1 & 0 & -7 & 6 \\ 2 & & 2 & 4 & -6 \\ \hline & 1 & 2 & -3 & 0 \end{array} $$ Thus we can say that $x^3-7x^2+6=(x-2)(x^2+2x-3)$. Similarly $$ \begin{array}{r|rr|r} & 3 & -8 & 4 \\ 2 & & 6 & -4\\ \hline & 3 & -2 & 0 \end{array} $$ so $3x^2-8x+4=(x-2)(3x-2)$.

Since you said you want other methods, you can use the substitution $t=x-2$, so the limit becomes $$ \lim_{t\to0}\frac{(t+2)^3-7(t+2)+6}{3(t+2)^2-8(t+2)+4} = \lim_{t\to0}\frac{t^3+6t^2+12t+8-7t-14+6}{3t^2+12t+12-8t-16+4} = \lim_{t\to0}\frac{t^3+6t^2+5t}{3t^2+4t} $$ Factor out $t$ and you're done.

Also the big gun (also known as l'Hôpital) works: $$ \lim_{x\to2}\frac{x^3-7x+6}{3x^2-8x+4}= \lim_{x\to2}\frac{3x^2-7}{6x-8} $$

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