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in the problem:

$$\lim_{u\to4}\frac{u^3-7u^2+12u}{4-u}$$

I keep getting the end result of $\frac{0}{0}$

I have tried things like:

$$\lim_{u\to4}\frac{u^3(-\frac{7u^3}{u^3}+\frac{12}{u^3})}{u^3(\frac{4}{u^3}-\frac{u}{u^3})}$$

But no no avail as I am still getting the end result of $\frac{0}{0}$

I have also tried plugging in numbers close to 4 and get a limit of $-4$. However when I plug this into wolframalpha it is telling me the limit is $4$ not $-4$.

Could anyone give me some insight on this?

Yiorgos S. Smyrlis
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Bob
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    The easiest approach is to factor $u-4$ from the numerator, and then simplify. – Andrés E. Caicedo Sep 29 '13 at 18:15
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    HINT: $u^3-7u^2+12u=u(u^2-7u+12)=u(u-4)(u-3)=-u(4-u)(u-3)$ – Mufasa Sep 29 '13 at 18:16
  • Limits of expressions in which the numerator and denominator both approach $0$ are the principal reason why the concept of limit is important. – Michael Hardy Sep 29 '13 at 18:34
  • By the way, you don't have to be good at factoring to handle limits like this. Since the numerator equals zero when you plug in $u=4,$ it follows from the Factor Theorem (covered in precalculus or high school Algebra 2) that $u-4$ is a factor of the numerator. Now just divide $u-4$ into the denominator (ordinary high school algebra division, or you can use synthetic division) to factor the numerator. For more about using the Factor Theorem to factor polynomials, see my answer at Finding limit of a quotient. – Dave L. Renfro Jan 27 '14 at 15:18

3 Answers3

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Factor! Numerator factors to $$u(u^2 - 7u + 12) = u(u-4)(u - 3)$$

Now negate the denominator, $$\dfrac{u(u-4)(u-3)}{-(u - 4)}$$ and cancel, assuming $u\neq 4$, so that we are not dividing by $u - 4 = 4-4=0$ to cancel.

$$\dfrac{u(u-4)(u-3)}{-(u - 4)} = -u(u-3) = 3u - u^2$$

NOW evaluate the limit. Recall, as $x$ approaches $4$, the limit of a function can exist even when the function is not defined at $4$.

amWhy
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  • how are you getting $-(u-4)$ in the denominator when the problem originally stated $(4-u)$? – Bob Sep 29 '13 at 18:25
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    I am saying that $$-(u - 4) = -1(u - 4) =(-1)u - (-1)\cdot 4 = - u + 4 = (4-u)$$ The negative distributes, since negating an expression is equivalent to multiplying the expression by $-1$. – amWhy Sep 29 '13 at 18:27
2

Observe that $u^3-7u^2+12u=u(u^2-7u+12)=u(u-3)(u-4)$

$$\implies \frac{u^3-7u^2+12u}{4-u}=\frac{u(u-3)(u-4)}{4-u}=-u(u-3)\text{ if }u-4\ne0$$

as $u\to4, u\ne4\iff u-4\ne0$

lab bhattacharjee
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2

The technique you tried works for limit at $\infty$ (think about what happens to the individual terms when $u \to \infty$ versus $u \to 0$ or $u \to 4$), which is why it isn't helping you here.

Try making a change of variable $u = x+4$, if limits at 0 are easier for you to grasp. Or even $u = 4 + 1/x$ if you have a better time with limits at infinity.

  • Thanks, I didn't realize that. This will help me out in the future so I don't waste time trying it. – Bob Sep 29 '13 at 18:39