I'm currently studying delta-epsilon limits and came across this problem that I can't understand the answer for. I've solved the problem multiple times and each time I get the same answer, shown below, that seems to make sense, but the given answer is completely different. The problem is thus: For each given $\epsilon$, find $\delta$ so that if $0 < |x-a| < \delta$ then $|f(x) - L| < \epsilon$
The limit is $\lim_{x \to 0} (x^2+3)=3$ Find the values of $\delta$ for the following three values of $\epsilon$:
- $\epsilon = 2$
- $\epsilon = 1$
- $\epsilon = 0.3$
These are the steps I used to solve it:
- By looking at the limit, $L$ is $3$ and $a$ is $0$.
- Pick the first value for $\epsilon = 2$
- The inequality is $|f(x) - L|<\epsilon$
- The inequality filled in is $|x^2+3-3|<2$
- Simplifying: $|x^2|<2$
- The target form to determine $\delta$ is $|x-a|<\delta$
- Thus $|x|<\delta$ as $a=0$
- Converting line 5 to the form of line 7: $\sqrt {|x^2|} < \sqrt 2$
- Thus $|x| < \sqrt 2$
- Thus, given lines 7 and 9, $\delta = \sqrt 2$
- Thus, the transform from a given $\epsilon$ to a given $\delta$ is $\delta=\sqrt \epsilon$
- Solving for all three given values yields the answers $\delta = \sqrt 2$, $\delta = \sqrt 1 = 1$, and $\delta = \sqrt {0.3}$
I checked these answers against the solutions manual and the answer I got was completely different. The book states $$\delta=\frac{1}{5}\epsilon$$ and thus the correct answers are $\delta=0.4$, $\delta=0.2$, and $\delta=0.06$.
Who's right? If I'm wrong, where did I make an error in my steps? It seems to me that it is impossible to go from $x^2$ to $\frac{1}{5}$, so is it possible that the book's answer is incorrect?