As usual, I'm having trouble, not with the calculus, but the algebra. I'm using Calculus, 9th ed. by Larson and Edwards, which is somewhat known for racing through examples with little explanation of the algebra for those of us who are rusty.
I'm trying to prove $$\lim_{x \to 1}(x^2+1)=2$$ but I get stuck when I get to $|f(x)-L| = |(x^2+1)-2| = |x^2-1| = |x+1||x-1|$. The solution I found says "We have, in the interval (0,2), |x+1|<3, so we choose $\delta=\frac{\epsilon}{3}$."
I'm not sure where the interval (0,2) comes from.
Incidentally, can anyone recommend any good supplemental material to go along with this book?
Best Answer
Because of the freedom in the choice of $\delta$, you can always assume $\delta < 1$, that implies you can assume $x$ belongs to the interval $(0, 2)$.
Edit: $L$ is the limit of $f(x)$ for $x$ approaching $x_0$, iff for every $\epsilon > 0$ it exists a $\delta_\epsilon > 0$ such that: $$\left\vert f(x) - L\right\vert < \epsilon$$ for each $x$ in the domain of $f$ satisfying $\left\vert x - x_0\right\vert < \delta_\epsilon$. Now if $\delta_\epsilon$ verifies the above condition, the same happens for each $\delta_\epsilon'$ such that $0 < \delta_\epsilon' < \delta_\epsilon$, therefore we can choose $\delta_\epsilon$ arbitrarily small, in particular lesser than 1.