Calculate $\lim_{x\rightarrow1}\frac{(1-x^2)}x$ by definition

Question

How do I calculate the $\lim_{x\rightarrow1}\frac{(1-x^2)}x$ using the δ/ε definition?

I'm losing my mind on this :c

I'm like lost after something like this:

|$\frac{(1+x) (1-x)}x$ - L| < ε

Answer 1

To get you started, if you want to show that

$$\lim_{x \to c} f(x) = L$$

then you show, for all $\varepsilon > 0$, there exists $\delta > 0$ such that

$$0< | x-c| < \delta \;\;\;\;\; \text{implies} \;\;\;\;\; | f(x) - L | < \varepsilon$$

You wish to show that, in that case,

$$\lim_{x \to 1} \frac{1-x^2}{x} = L$$

But you need to first determine what $L$ should be. Looking at a graph would give us a good start: $L=0$.

To confirm this we want to show that, for all $\varepsilon > 0$, there exists some $\delta > 0$ such that

$$0 < |x-1| < \delta \;\;\;\;\; \text{implies} \;\;\;\;\; \left| \frac{1-x^2}{x} \right| < \varepsilon$$

Your goal: find the necessary $\delta$.

Wikipedia also has some worked examples you might find useful. I think that the thing you'll find common in these is that the $L$ you're looking for is "given." To my understanding you can't really find the limit outright from the definition, just verify whether something is indeed the limit. In that sense, then, you need to figure out the limit $L$ should be and then verify it by finding the corresponding $\delta$.

Some more definitions and examples can be found on Paul's Online Math Notes and on Brilliant.

Answer 2

Given $\varepsilon$, when $0<|1-x|<\delta < \frac{\varepsilon}{6}$, then $$ \bigg|\frac{1-x^2}{x}-0\bigg|= \bigg| \frac{(1-x)(1+x)}{x}\bigg| <\delta \frac{2+2\delta}{1-\delta }<6\delta< \varepsilon$$

Answer 3

$\lim_\limits{x\to 1} \frac {1-x^2}{x} = 0$

We must show that for any $\epsilon > 0$ there is a $\delta >0$ such when $|x-1|<\delta$ it implies that $|\frac {1-x^2}{x} | < \epsilon$

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

Let's insist that $\delta < \frac 12$ then $|\frac {1+x}{x}| < 5$ (i.e. $\frac 12 < x < \frac 32 \implies x+1 < \frac 52$ and $x > \frac 12)$

$|\frac {(1-x^2)}{x}| < 5\delta < \epsilon$

$\delta = \min (\frac 12, \frac {\epsilon}{5})$

Answer 4

Let $$0<|1-x|<\delta$$ $$ \Rightarrow |(1-x) -0| <\delta \Rightarrow \left|(1-x) \frac{2-\delta}{1-\delta}-0\right| <\delta \left|\frac{2-\delta}{1-\delta}\right| $$ $$\Rightarrow \left|\frac{(1-x)(1+x)}{x}-0\right|<\delta \left|\frac{2-\delta}{1-\delta}\right|$$ $$ \Rightarrow \left |\frac{1-x^2}{x}-0 \right|< \epsilon(\delta) \Rightarrow \lim_{x\rightarrow 1} \frac{1-x^2}{x} \Rightarrow 0.$$