Formal definition of limit

Question

Use the formal definition of limit to verify the indicated limit:

$$\lim\limits_{x\to\frac{1}{2}}\frac{1-4x^2}{1-2x} = 2$$

anyone know how to go about this problem?

Answer 1

Hint: $a^2 - b^2 = (a+b)(a-b)$

$$ \epsilon > \left|\frac{1-4x^2}{1-2x} - 2 \right| = |2x -1| = \left|2\left(x - \frac{1}{2}\right)\right| = 2\left|x-\frac{1}{2}\right| $$

Choose $\delta = \epsilon / 2$.

Answer 2

Hint: you need to show that for every $\epsilon>0$ there exists a $\delta>0$ such that

$$0<\left|x-\frac{1}{2}\right|<\delta\quad\implies\quad\left|\frac{1-4x^2}{1-2x}-2\right|<\epsilon.$$

Answer 3

Let $\epsilon >o$. Note that

$$\dfrac{1-4x^2}{1-2x}-2=\dfrac{-4x^2+4x-1}{1-2x}=\dfrac{-(1-2x)^2}{1-2x}=-1+2x$$. So, If you take $\delta <\dfrac{\epsilon}{2}$, you get your answer. Here we are allowed cancel $1-2x$ both numerator and denominator since we are sure that $x\neq \frac{1}{2}$ since we are dealing weed limits.

Answer 4

You know that

$$\dfrac{1-4x^2}{1-2x}-2=2x-1$$

It is assumed that the limit of $2x-1$ for $x \rightarrow \frac{1}{2}$ is $0$.

From the definition of limit, yoy have that:

$$\forall\epsilon >0 ~\exists \delta > 0: \left|x - \frac{1}{2}\right|<\delta \Rightarrow \left|2x-1 - 0\right| < \epsilon$$

This means that:

$$\left|2x-1 - 0\right| < \epsilon \Rightarrow -\epsilon<2x - 1 < \epsilon$$

which simplifies into:

$$\frac{1 - \epsilon}{2} < x < \frac{1 + \epsilon}{2}$$

If you impose that $\delta = \frac{\epsilon}{2}$, then:

$$-\delta < x - \frac{1}{2} < \delta$$

which is $\left|x - \frac{1}{2}\right| < \delta$.

Finally,

$$\left|x - \frac{1}{2}\right| < \delta \Rightarrow \frac{1 - \epsilon}{2} < x < \frac{1 + \epsilon}{2} \Rightarrow \left|2x-1 - 0\right| < \epsilon$$

where for each $\epsilon > 0$ exists a $\delta = \frac{\epsilon}{2} > 0$

Answer 5

I suggest you to use the fact that $a^2 - b^2 = (a+b)(a-b)$ and then you end up with a polynomial of degree 1 in which the definition of limit is easier to apply.