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Proving that $$\lim_{x\to 4}\frac{2x-5}{6-3x} = -\frac{1}{2} $$

Now of course when you plug in $4$ you get $-\frac{1}{2}$, but I need to prove it using:

$\forall \varepsilon \gt 0,\;\exists\delta \gt 0\;\forall x \in D $ (where $D$ is the domain of the function)

$$0\lt |x-x_0|\lt \delta \implies |f(x)-L| \lt \varepsilon $$

Where $L$ is the limit we "guessed" and $x_0$ is where $x$ approaches)

So far I have:

$$\begin{align} 0\lt |x-4|\lt \delta &\implies \left|f(x)-\left(-\frac{1}{2}\right)\right| \lt \varepsilon\\ \iff 0\lt |x-4|\lt \delta &\implies \left|\frac{2x-5}{6-3x}+\left(\frac{1}{2}\right)\right| \lt \varepsilon\\ \iff 0\lt |x-4|\lt \delta &\implies \left|\frac{4-x}{6(x-2)}\right| \lt \varepsilon \end{align}$$

This is where I'm stuck. What exactly do I need to prove after this last line?

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  • $\begingroup$ Change the $4-x$ in the num into $x-4$ which then can be divided by $x-4$ in front of your delta. Now you can estimate epsilon as $x$ goes to 4 $\endgroup$ – imranfat Mar 19 '16 at 18:07
  • $\begingroup$ Now you have to prove that for any choice of epsilon, you can find a delta where this is true. $\endgroup$ – fleablood Mar 19 '16 at 18:07
  • $\begingroup$ @imranfat not really following what you mean, any chance you could show me? $\endgroup$ – user300011 Mar 19 '16 at 18:12
  • $\begingroup$ The edit of Subhadeep has made things clearer. Now when you divide both sides by $|x-4|$, you have an expression for delta. What happens if you plug in $x=4$? $\endgroup$ – imranfat Mar 19 '16 at 18:37
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Given any $\varepsilon > 0$, let $\delta = \min\{1, \varepsilon/M\} > 0$, where $M$ is some fixed magical positive constant. While doing this proof, we will eventually figure out what $M$ should be, then go back and erase $M$ and replace it with this constant. Now if $0 < |x - 4| < \delta$, then observe that: \begin{align*} \left| \frac{2x - 5}{6 - 3x} - \frac{-1}{2} \right| &= \left| \frac{4 - x}{6(x - 2)} \right| \\ &= \frac{1}{6|x - 2|}|x - 4| \\ &< \frac{1}{6|x - 2|} \cdot \frac{\varepsilon}{M} &\text{since } |x - 4| < \delta \leq \frac{\varepsilon}{M} \\ \end{align*} Now we take a break from our proof and try to figure out a way to bound $\frac{1}{6|x - 2|}$ by a constant $M$ given that $x$ is at most $1$ away from $4$, implying that $|x - 2|$ won't be too close to zero. Indeed, if $|x - 4| < 1$, then observe that: \begin{align*} |x - 2| &= |(x - 4) - (-2)| \\ &\geq ||x - 4| - \left| -2 \right|| &\text{by the reverse triangle inequality} \\ &= 2 - |x - 4| &\text{since } |x - 4| < 1 \leq 2 \\ &> 2 - 1 &\text{since } |x - 4| < 1 \\ &= 1 \end{align*} so that: $$ \frac{1}{6|x - 2|} < \frac{1}{6} $$ Thus, by taking $M = \frac{1}{6}$, we can finish up our original proof:


Given any $\varepsilon > 0$, let $\delta = \min\{1, 6\varepsilon\} > 0$. Then if $0 < |x - 4| < \delta$, observe that: \begin{align*} \left| \frac{2x - 5}{6 - 3x} - \frac{-1}{2} \right| &= \left| \frac{4 - x}{6(x - 2)} \right| \\ &= \frac{1}{6|x - 2|}|x - 4| \\ &< \frac{1}{6|x - 2|} \cdot 6\varepsilon &\text{since } |x - 4| < \delta \leq 6\varepsilon \\ &< \frac{1}{6} \cdot 6\varepsilon &\text{since } |x - 4| < \delta \leq 1 \\ &= \varepsilon \end{align*} as desired. $~\blacksquare$

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  • $\begingroup$ Thanks for the response @Adriano! Any chance you could explain the step with $\delta= min (1, \frac{\epsilon}{M})$? $\endgroup$ – user300011 Mar 21 '16 at 18:46
  • $\begingroup$ Can't edit previous comment, Why did we pick 1 and where do we get the $\frac{\epsilon}{M}$? $\endgroup$ – user300011 Mar 21 '16 at 18:54
  • $\begingroup$ The $1$ is arbitrary. We can pick any fixed constant; it's just a nice enough round number to work with. Its purpose is to bound the stuff other than $|x - 4|$ by some number, say $M$. This explains the $\varepsilon/M$, since multiplying $M$ by $\varepsilon/M$ will yield the desired $\varepsilon$ bound. $\endgroup$ – Adriano Mar 21 '16 at 19:19
  • $\begingroup$ Hm. For other proofs would it be wise using the 1 again as a starting point? Is it something I can always use? $\endgroup$ – user300011 Mar 21 '16 at 19:31

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