Prove the limit l for the function at a:
$$f(x)= x^4 + \frac1x, a =1$$
I have successfully found a $\delta$ in terms of $\epsilon$, and here is how I did it:
Since we can see the limit is 2 at a = 1, the definition of a limit looks like:
$$|x^4 + \frac1x - 2|< \epsilon, \text{ then } 0<|x-1|< \delta$$
$$|x^4 + \frac1x - 2| = |x-1| |\frac1x| |x^4+x^3+x^2+x-1|$$
Let's restrict $|x-1|< \frac12$ , so that now $\frac12 < x < \frac32$, which means $|\frac1x| < 2$.
Similarly, $|x^4+x^3+x^2+x-1| < 22.375$
Therefore, $|x-1| |\frac1x| |x^4+x^3+x^2+x-1| < 2(22.375)|x-1| < \epsilon$, so $|x-1| < \frac{8\epsilon}{179}$
Hence $\delta = \min (\frac12, \frac{8\epsilon}{179}) \Box$
But this is not how Spivak did it, and I am absolutely bamboozled. Spivak's answer in the back of the book looks like this:
$$|\frac1x - 1| < \epsilon \text{ for } 0<|x-1|<\delta_1=\min(\frac12,\frac\epsilon2)$$
So that
$$|\frac1x - 1| < \frac\epsilon2 \text{ for } 0<|x-1|<\bar{\delta_1}=\min(\frac12,\frac\epsilon4)$$
Similarly, the solution to the previous problem (which I understand no problem) gives a $\delta_2$ such that
$$|x^4-1|<\epsilon \text{ for } 0<|x-1|<\delta_2$$
and we have a corresponding $\bar{\delta_2}$. Then we can take $\delta=\min(\bar{\delta_1},\bar{\delta_2}) \Box $
I understand the first line, but am totally lost on the second. Why is he then setting $|\frac1x-1|< \frac\epsilon2$? And how can he say $|x^4-1|< \epsilon$, when we have $|x^4+x^3+x^2+x-1|$? Any insight on this would be greatly appreciated!
The answer in the book is tricky because it divides the problem into two parts first he proved that there exists a $\overline{\delta_1}$ such that : $$|\frac1x - 1| < \frac\epsilon2 \text{ for } 0<|x-1|<\overline{\delta_1}=\min(\frac12,\frac\epsilon2) $$ and the same argument gives you a $\overline{\delta_2}$ such that: $$|x^4-1|<\color{#a00}{\frac{\epsilon}{2}} \text{ for } 0<|x-1|<\overline{\delta_2} $$ And as a final result for $|x-1|<\delta=\min(\overline{\delta_1},\overline{\delta_2})$ we have : $$\left|x^4+\frac{1}{x}-2\right|\leq |x^4-1|+ |\frac1x - 1| <\frac\epsilon2+\frac\epsilon2=\epsilon $$