Using the precise definition of limits to prove a limit

Question

Question

Let $a$ be a positive real number. Use the $\epsilon-\delta$ definition of limits to prove that $$\lim\limits_{x\to a}\ \frac {3x^2 + 2ax + a^2} {2x + a} = 2a\ .$$

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My working

Let $\epsilon > 0$ be given.

For $|x - a| < \delta$, choose $\delta = \frac 1 3 \epsilon - 2a$.

\begin{align} |x + a| & = |x - a + 2a| \\[5 mm] & \leq |x - a| + 2a \\[5 mm] & < \frac 1 3 \epsilon \end{align}

\begin{align} \left|\frac {3x^2 + 2ax + a^2} {2x + a} - 2a\right| & = \left|\frac {3x^2 - 2ax - a^2} {2x + a}\right| \\[5 mm] & = \left|\frac {(3x + a)(x - a)} {2x + a}\right| \\[5 mm] & < \left|\frac {3(x + a)(x - a)} {2(x - a)}\right| \\[5 mm] & = \frac 3 2 |x + a| \\[5 mm] & < \frac 1 2 \epsilon \\[5 mm] & < \epsilon \end{align}

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May I know if I have chosen a suitable $\delta$ and if my proof is logical? Any comments/suggestions to improve my working will be greatly appreciated :)

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Edit

Thank you everyone for your inputs! I have learnt a lot :)

Answer 1

Your $\delta$ doesn't work and it would be strange if it did, because in almost every case $\delta$ must be very close to $0$ when $\varepsilon$ is very close to $0$. But your $\delta$ is always greater than $2a$.

Note that$$\frac{3x^2+2ax+a^2}{2x+a}-2a=\frac{(3x+a)(x-a)}{2x+a}$$and that therefore$$\left|\frac{3x^2+2ax+a^2}{2x+a}-2a\right|=\left|\frac{3x+a}{2x+a}\right||x-a|.\tag1$$If $|x-a|<\frac a4$, then $|x|<\frac54a$ and so $|3x+a|<\frac{19}4a$. On the other hand,\begin{align}|2x+a|&=|2(x-a)+3a|\\&\geqslant3a-2|x-a|\\&>\frac52a.\end{align}So,$$\left|\frac{3x+a}{2x+a}\right|<\frac{19}{10}.$$Therefore, it follows from $(1)$ that $\delta=\min\left\{\frac a4,\frac{10}{19}\varepsilon\right\}$ will work.

Answer 2

There are at least two major errors.

The first is where you try to choose $\delta$ in such a way to ensure that $$|x - a| + 2a \stackrel?< \frac13 \epsilon. \tag1$$ Leaving aside the problems with your method of choosing $\delta,$ just take a good long look at Inequation $(1)$.

Now recall that $a$ is some given positive integer and that absolute values are never negative. Therefore the left hand side is never less than the positive number $2a.$ So if we just consider the case $\epsilon = a,$ it should be clear that $\frac13 \epsilon = \frac13 a < 2a \leq |x - a| + 2a,$ and therefor Inequation $(1)$ is false for this particular choice of a positive number $\epsilon$. Now do you recall that part of the definition of a limit where it says "for all $\epsilon > 0$"? It means that whatever you do, you have to have an answer for every positive $\epsilon.$ If there is even one positive $\epsilon$ for which your proof doesn't work, your proof doesn't work. And that's how it will go if you try to base a proof on Inequation $(1)$.

The second major error I found is here: $$ \left|\frac {(3x + a)(x - a)}{2x + a}\right| \stackrel?< \left|\frac {3(x + a)(x - a)}{2(x - a)}\right|. \tag2 $$

Note that the right-hand side is not even defined unless $x - a\neq 0,$ but in the context of the definition of a limit we actually have $0 < |x-a| < \delta$, not just $|x-a| < \delta$, so that's OK. And given that $x - a \neq 0,$ we can cancel a factor of $|x - a|$ on both sides, which means you actually need this to be true: $$ \left|\frac {3x + a}{2x + a}\right| \stackrel?< \left|\frac {3(x + a)}{2(x - a)}\right|. $$

So let $x - a = t$, where $0 < |t| < \delta.$ Then $x = a + t$ and $$ \left|\frac {3x + a}{2x + a}\right| = \left|\frac {4a+3t}{3a + 2t}\right|. $$ For very small values of $t$ the right-hand side is approximately $\frac 43.$

On the other hand, $$ \left|\frac {3(x + a)}{2(x - a)}\right| = \left|\frac {6a + 3t}{2t}\right|, $$ which gets very large as $t$ gets very small. For example, if $0 < t < \frac1{100}a,$ the right-hand side is greater than $200.$ So in order to show that Inequation $(2)$ is true, you need to show that at least some numbers greater than $200$ are approximately equal to $\frac 43.$ That is not true even for the largest possible value of $\frac43,$ so a proof that relies on Inequation $(2)$ won't work.

Answer 3

Often a substitution can simplify the calculations or clarify the issues. Let $f(x)=\frac {3x^2+2ax+a^2} {2x+a}$ when $2x+a\ne 0.$ Let $x=y+a.$ When $2x+a\ne 0$ we have $$|f(x)-2a|=|y|\cdot\frac {|a+3y|}{|3a+2y|}.$$ Now if $|y|<|a|$ then (i)... $|a+3y|\le |a|+3|y|<4|a|=4a\;$ and (ii)... $|3a+2y|\ge |3a|-|2y|>|a|=a>0,\;$ so (iii)... $\frac {|a+3y|}{|3a+2y|}<4,\;$ so $$|f(x)-2a|\le 4|y|.$$ So, given $\epsilon>0,$ let $\delta=\min(\epsilon/4, a).$ Now if $|x-a|<\delta$ then $f(x)$ exists (because if $|x-a|<a>0$ then $2x+a\ne 0$) and $$|f(x)-2a|\le 4|y|=4|x-a|<4\delta\le\epsilon.$$