Using $\epsilon-\delta$ to show that $\lim \frac{x^2 + 1}{2x^3 - 1}$ as $x \rightarrow 1$ is 2.

Question

So this looks stupid easy, and i'm pretty sure it is. But I have been working on it (among other problems) all day and cannot figure out. My first approach was to use the limit property to break it up into two separate limits and then just show that the equality held, but my tasks failed when I couldn't do the proof for $\frac{1}{2x^3-1}$. I wasn't really shown any process on how to do this and i'm pretty lost. I've gathered from some introductory videos on youtube that I want to play with $x$ and $\epsilon$ until I have a expression of $\epsilon$ that I can set my $\delta$ equal to.

To start off, we want to find $0<|x - 1| < \delta$ that implies $|\frac{x^2 + 1}{2x^3 - 1} - 2| = | \frac{-4x^3 + x^2 + 2}{2x^3 - 1}| < \epsilon$ for any given $\epsilon$. This is just straight from the definition. But I don't have any clever tricks to proceed.

Any help is greatly appreciated.

Answer 1

We must show that $$\left|\frac{x^2 + 1}{2x^3 - 1} - 2\right| = |x-1| \cdot \left|\frac{4x^2 + 3x + 3}{2x^3 - 1}\right| < \epsilon$$

This expression was derived from the factor theorem and polynomial division. We need an upper bound on $\left|\frac{4x^2 + 3x + 3}{2x^3 - 1}\right|$. Restrict $|x-1| < \frac{1}{10}$. The numerator is bounded above in $(0.9, 1.1)$ by $100$ and the denominator is bounded below by $0.1$. This is not difficult to show rigorously, since the numerator and denominator are monotonically increasing in the interval. Also note that both the numerator and denominator are positive.

Thus we have $$\left|\frac{x^2 + 1}{2x^3 - 1} - 2\right| = |x-1| \cdot \left|\frac{4x^2 + 3x + 3}{2x^3 - 1}\right| < |x-1| \cdot \left|\frac{100}{0.1}\right|< \epsilon$$

We can thus choose $\delta < \min{(\frac {1}{10}, \frac{\epsilon}{1000})}$

Answer 2

If $f(x) =\frac{x^2+1}{2x^3-1} $, to see what happens as $x \to 1$, I apply my saying "always have limits that go to zero."

Therefore, let $x = 1+y $, so we can let $y \to 0$.

Then $f(x) =\frac{x^2+1}{2x^3-1} =\frac{(1+y)^2+1}{2(1+y)^3-1} =\frac{2+2y+y^2}{1+3y+3y^2+y^3} $, and now it is quite obvious that this goes to $2$ as $y \to 0$.