How to use $\epsilon-\delta$ definition to prove $\lim_{x\to1}\frac{x^3-3x+2}{x^4-4x+3}=\frac{1}{2}$?

Question

How to use $\epsilon-\delta$ definition to prove $\lim_{x\to1}\frac{x^3-3x+2}{x^4-4x+3}=\frac{1}{2}$?

My attempt:

$|\frac{x^3-3x+2}{x^4-4x+3}-\frac{1}{2}|=|\frac{(x+1)(x-1)}{2(x^2+2x+3)}|$ by some factorization.

I am confused to choose a $\delta>0$ such that for any $\epsilon>0$, if $0<|x-1|<\delta$ then $|\frac{x^3-3x+2}{x^4-4x+3}-\frac{1}{2}|<\epsilon$.

Could anyone help me? Thank you so much.

Answer 1

You are actually really close to completing the proof. Note that

$$\left|\frac{(x+1)(x-1)}{2(x^2+2x+3)}\right|\leq \left|\frac{x+1}{2(x^2+2x+3)}\right||x-1|$$

Now, the $|x-1|$ piece is less than $\delta$ (whatever you want to choose for $\delta$). However, based on our choice of $\delta$, we can bound

$$\left|\frac{x+1}{2(x^2+2x+3)}\right|$$

For example, if $\delta\leq \frac12$, then $\frac12\leq x\leq\frac32$. This implies

$$\left|\frac{x+1}{2(x^2+2x+3)}\right|=\frac{x+1}{2(x^2+2x+3)}\leq \frac{(3/2)+1}{2((1/2)^2+2(1/2)+3)}=\frac{5}{17}$$

In fact, this test $\delta$ will work great. Simply set

$$\delta=\text{min}\left\{\frac{1}{2},\epsilon\right\}$$

If $\epsilon>\frac12$ then $\delta=\frac12$ and

$$\left|\frac{(x+1)(x-1)}{2(x^2+2x+3)}\right|\leq\left|\frac{x+1}{2(x^2+2x+3)}\right||x-1|<\frac{5}{17}\delta=\frac{5}{17}\cdot\frac{1}{2}=\frac{5}{34}<\frac{1}{2}<\epsilon$$

If $\epsilon\leq \frac12$ then $\delta=\epsilon$ which gives

$$\left|\frac{(x+1)(x-1)}{2(x^2+2x+3)}\right|\leq\left|\frac{x+1}{2(x^2+2x+3)}\right||x-1|<\frac{5}{17}\delta=\frac{5}{17}\epsilon<\epsilon$$

The key here is working backwards and trying to find a 'test' $\delta$ which will make the function in the absolute values less than $\epsilon$. That is, once you set your 'test' $\delta$ you can get bounds on $x$ and then plug those bounds into your function. You'll notice that none of the bounds above are particularly tight. Thats because I tested $\delta \leq \frac12$ first and it worked. It's possible (probable), that there is a better bound greater than $\frac12$ which also works for delta. In the same vein, anything less than $\frac12$ would also work. The key is testing a $\delta$ small enough that is easy to work with in order to simplify the expressions.

Answer 2

Let $|x-1| <1/2$, then $x>0$:

$|\dfrac{(x+1)(x-1)}{2(x^2+2x+3)}| \le$

$\dfrac{(x+1)|(x-1)|}{2(2x+2)} =$

$\dfrac{(x+1)|(x-1)|}{4(x+1)} \le |x-1|$;

Let $\epsilon >0$ be given.

Choose $\delta =\min (1/2, \epsilon).$

Then $|x-1|<\delta$ implies

$|\dfrac{(x-1)(x+1)}{2(x^2+2x+3)}| \le |x-1|<\delta \le \epsilon$.