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Prove the following limits using only the epsilon delta definition:

Q1:

$$\lim_{x\to2^-} \sqrt{4-x^2}= 0$$

and

Q2:

$$\lim_{x\to\infty}\dfrac{x^2+2x}{x^2+1} = 1$$

For 1, I got stuck at the part whereby you obtained square root of $(2-x)$ and square root of $(2+x)$. Then can't continue, and not sure of what delta to choose.

For 2, not sure how to do it. I mean, if I obtained absolute of $\dfrac{x^2+2x}{x^2+1}$, but I am still stuck and not sure how to carry on. Not sure what delta to choose as well.

Mhenni Benghorbal
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troubledouble
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2 Answers2

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For question $1$, given any $\epsilon > 0$, your goal is to find a $\delta > 0$, such that for all $x \in (2- \delta, 2)$, we get that $\sqrt{4-x^2} < \epsilon$.

Now lets do reverse engineering. If we want $\sqrt{4-x^2} < \epsilon$, then we need $4-x^2 < \epsilon^2$, i.e., we need $x^2 > 4-\epsilon^2$. Hence, we need to choose $\delta$ such that $$(2-\delta)^2 > 4-\epsilon^2 \implies 2-\delta > \sqrt{4-\epsilon^2} \implies \delta < 2- \sqrt{4-\epsilon^2}$$

Hence, given any $\epsilon > 0$, if we choose $\delta = 2-\sqrt{4-\epsilon^2}$, we then have that for all $x \in (2-\delta,2)$, i.e., for all $x \in (\sqrt{4-\epsilon^2},2)$, we have $\sqrt{4-x^2} < \epsilon$.

For the second one, note that for large enough $x$, i.e., for $x > \dfrac12$, we have $$1 < \dfrac{x^2+2x}{x^2+1} < \dfrac{x^2+2x}{x^2} = 1 + \dfrac2x$$ Now given an $\epsilon$, can you make a choice of appropriate $\delta$ to justify that the limit is indeed $1$?

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Hints:
Q1. If $2-\delta < x < 2$ then $(2-x)(2+x) < 4 \delta$.

Q2. If $x>\delta$ then $\dfrac{x^2+2x}{x^2 + 1}-1=\dfrac{2x-1}{x^2 + 1}<\dfrac2x<\dfrac{2}{\delta}$.

njguliyev
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  • Sorry, I still don't understand how to do.... the hints are a little vague. For question 1 and 2, is there another way other than the reverse engineering method? – troubledouble Sep 15 '13 at 06:04