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I was wondering how get the proof of this limit:

$$\lim\limits_{x\to -\infty}\dfrac{{x^2} - x + 1}{x + 4} = -\infty$$

The problem is that I don't know what to do for find the appropriated values to make valid the implication of the formal definition (epsilon-delta).

I would appreciate if somebody can help me.

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    $\begingroup$ A possibility: as Kay K. notes, $$\frac{{x^2} - x + 1}{x + 4} = x-5+\frac{21}{x+4} $$ so it is sufficient to show that $\frac{21}{x+4}$ goes to $0$ using the $(\varepsilon,\delta)$-definition of limits (proving that $x-5 \to -\infty$ by the same arguments will be trivial). $\endgroup$ – Clement C. Dec 12 '15 at 23:54
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Expand what you know (as one can tell from your comment below Kay K.'s deleted answer) to a "snapshot" argument is not difficult, in fact.

If $x < -4$, then $$ \frac{x^{2}-x+1}{x+4} = x-5 + \frac{21}{x+4} < -9 + \frac{21}{x+4}; $$ given any $M < -9$, we have $-9 + 21/(x+4) < M$ if in addition $$ x < \frac{21}{M+9} - 4. $$

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You want to find $N(M)<0$ (a function in terms of $M<0$) such that

$$x<N(M)\implies \frac{x^2-x+1}{x+4}<M$$

Let $N(M)\le -4$. Then $x<-4$ and $$\frac{x^2-x+1}{x+4}<M\iff x^2-x+1>M(x+4)$$

$$\iff x^2-x(M+1)+(1-4M)>0$$

If $M\le -9-2\sqrt{21}$, then $\Delta=M^2+18M-3\ge 0$ and let $$N(M)=\min\left\{-4,\frac{M+1-\sqrt{M^2+18M-3}}{2}\right\}$$

If $M\in(-9-2\sqrt{21},0)$, then let $$N(M)=\min\left\{-4,\frac{k+1-\sqrt{k^2+18k-3}}{2}\right\}$$

for any $k\le-9-2\sqrt{21}$, e.g. you can let $k=-19$:

$$N(M)=\min\left\{-4,-11\right\}=-11$$

Answer: you can let $$N(M)=\begin{cases}\min\left\{-4,\frac{M+1-\sqrt{M^2+18M-3}}{2}\right\}, && M\le -9-2\sqrt{21}\\-11, && M\in(-9-2\sqrt{21},0)\end{cases}$$

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Note that $$\frac{x^2-x+1}{x+4}=\frac{(x+4)(x-5)+21}{x+4}.$$ You need to show that given $M<0$, there exists $N<0$ such that $\frac{x^2-x+1}{x+4}<M$ for all $x<N$. Let $N=\min\{M,-4\}.$ Then if $x<N$, \begin{align*} \frac{x^2-x+1}{x+4}&=\frac{(x+4)(x-5)+21}{x+4}\\ &=(x-5) + \frac{21}{x+4}\\ &<(N-5) + \frac{21}{x+4}\\ &<M+\frac{21}{x+4}\\ &<M. \end{align*} We can drop the $\frac{21}{x+4}$ because $x<N \leq -4$, so $x+4$ is negative, so $\frac{21}{x+4}<0$.

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  • $\begingroup$ How you choose the value of N? $\endgroup$ – egarro Dec 13 '15 at 0:03
  • $\begingroup$ Why not make N <= M - 21/x+4 + 5 ? $\endgroup$ – egarro Dec 13 '15 at 0:12
  • $\begingroup$ It is x+4, not x-4, look the original fraction $\endgroup$ – egarro Dec 13 '15 at 0:25
  • $\begingroup$ My bad about the $x+4$; I've corrected it. You can't choose $N \leq M-\frac{21}{x+4}$ or anything involving $x$ because later you need to say "let $x<N$", and that might be inconsistent with how you've just specified $N$. The way I chose $N$ was to make the inequalities work out - we know $x-5<N-5$, so we need to make sure $N-5<M$. (So in fact in my edit, I could've made $N=M+5$, but it works as is.) $\endgroup$ – kccu Dec 13 '15 at 2:32

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