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I believe that I completed this problem correctly but I could use a second set of eye's to verify that I used the right methods. Also if you have a suggestion for a better method of how to solve this I would appreciate any advice.

Prove $$\lim_{x \to -1} x^{3}-2x=1$$ Given $\epsilon > 0,$ let ; $\delta= \min\left \{{1\over2}, {4\epsilon\over11} \right \} $

If $\; 0<\left | x+1 \right | \Rightarrow \delta \; then \; {-1\over4} < \left ( x-{1\over2} \right )^{2}-{5\over4} < {11\over4} $

so $\left | x^{3} - 2x - 1 \right | = \left | x+1 \right |\left | x^{2}-x-1 \right |=\left | x+1 \right |\left | \left ( x-{1\over2} \right )^{2}-{5\over4} \right |<{11\delta\over4}=\epsilon $

scratch work, let $\delta<{1\over2} \Rightarrow \left | x+1\right | < \delta < {1\over2} \Rightarrow{-1\over2} < x+1 < {1\over2} \Rightarrow -2 < x-{1\over2} < -1 \Rightarrow 4 > \left ( x -{1\over2} \right )^{2} > 1 \Rightarrow {11\over4} > \left ( x-{1\over2}\right )^{2}-{5\over4} > {-1\over4}$

Mahdi Khosravi
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Brooks
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Your method looks like taken "out of the blue": why did you choose $\;\delta\,$ as you did? Why did you do that odd-looking calculations in the third line (which I didn't understand right away, btw)?.

I propose the following: for an arbitrary $\,\epsilon>0\,$:

$$|x^3-2x-1|=|(x+1)(x^2-x-1)|<\epsilon\iff |x+1|<\frac\epsilon{|x^2-x-1)|}$$

Now the estimmation "trick": for $\,x\,$ "pretty close to $\,-1\,$ , we get $\,x^2-x-1\,$ "pretty close" to $\,1\,$ (you can either use freely this or formally prove by limits that $\,x^2-x-1\xrightarrow[x\to -1]{}1\;$).

Thus, we can choose, for example

$$\delta:=\epsilon+0.1\;,\;\;\text{so whenever}\;\;|x^2-x-1|>\frac{10\epsilon}{10\epsilon+1} $$ we'll get that

$$|x+1|<\delta\implies |(x^3-2x)-1|<\epsilon$$

DonAntonio
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$\lim_{x \to a} f(x) = f(a)$ iff $f$ is continuous at $a$. $x^3 - 2x$ is continuous since identity and constant functions are obviously continuous and sum, difference, and product of countinuous functions is continuous. So any polynomial is continuous and any limit of polynomial is simply function value.

user87690
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