I've been asked to show using $\epsilon – \delta$ definition of limit that $$\lim_{x\to 1} -x^4 = -1$$
To do so I've been asked that what is the value of $\delta$ would be from one of these two options :
a) $\min \{1,\frac{\epsilon}{15}\}$
b) $\frac{\epsilon}{15}$
I'm really confused as there's no other information given. And there's no way for me to ask for any more information.
What I have done so far is :
Let $\epsilon \gt 0$ be given, then
$$ \mid -x^4 – (-1) \mid \\
= \mid x^4 -1 \mid \\
= |x-1||x+1||x^2+1| \lt \epsilon \\ \implies |x-1| < \frac{\epsilon}{|x+1||x^2+1|}
$$
Now since the domain of the function is not given here and even though I know that $x\to 1$ but I am not being able to "generally" deduce the maximum value of the expression $\frac{1}{|x+1||x^2+1|}$. This is where I'm stuck.
Can anyone please help?? It's really urgent. Thank you.
Best Answer
Note that\begin{align}|-x^4-(-1)|&=|x^4-1|\\&=|x-1|\times|x^3+x^2+x+1|.\end{align}And, if $|x-1|<1$, then$$|x|=|(x-1)+1|\leqslant|x-1|+1<2,$$and therefore$$|x^3+x^2+x+1|\leqslant2^3+2^2+2+1=15,$$which implies that $|-x^4-(-1)|<15|x-1|$. So, if $|x-1|<\frac\varepsilon{15}$, then $|-x^4-(-1)|<\varepsilon$. In particular, $\delta=\min\left\{1,\frac\varepsilon{15}\right\}$ will work.