Epsilon delta definition: $\lim _{x\to-2} (2x^2+5x+3)=1$
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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jol
Updated on August 01, 2022Comments
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jol over 1 year
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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Mhenni Benghorbal about 11 yearsSee here.
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jol about 11 yearsthanks; so delta= minimum(epsilon/5,1)
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N. S. about 11 yearsYes. Keep in mind this trick, if you can make $\delta$ a common factor, you only need a very rough estimate for the rest.