# Epsilon delta definition: $\lim _{x\to-2} (2x^2+5x+3)=1$

I'm trying to use the epsilon delta definition to prove that $$\lim _{x\to-2} (2x^2+5x+3)=1$$

evaluating: $|(2x^2+5x+3)-1|\lt \epsilon$

under the condition: $0\lt |x-(-2)|\lt\delta$

I arrived at: $|((x+2)+(x+2)-3)(x+2)|\lt \epsilon$; which simplifies to: $(2\delta-3)(\delta)<\epsilon$

What to do now? Do I evaluate the prior expression so as to get an appropriate range and relation between epsilon/ delta, upon which the limit is condition. If so how?

btw, this question makes use of a similar previous question Use the epsilon-delta definition to prove the following statement.

Thanks

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See here. – Mhenni Benghorbal Oct 15 '12 at 3:04

First, your last conclusion should be

$$|2 \delta -3| \delta < \epsilon \,$$

since the bracket is probably negative.

Now you could solve the quadratic inequality, or simply estimate the LHS. Remember you need a $\delta$ which works, not the best $\delta$.

So, we can look for some $\delta <1$. Then

$$|2 \delta -3| \delta \leq (2 \delta +3) \delta =2 \delta^2+3\delta < 2 \delta +3 \delta = 5\delta$$

So, if you make $5 \delta < \epsilon$, you are done.. But don't forget that the argument works only if $\delta <1$.

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thanks; so delta= minimum(epsilon/5,1) – jol Oct 15 '12 at 3:01
Yes. Keep in mind this trick, if you can make $\delta$ a common factor, you only need a very rough estimate for the rest. – N. S. Oct 15 '12 at 3:34