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Numerade Educator



Problem 25 Easy Difficulty

Prove the statement using the $ \varepsilon $, $ \delta $ definition of a limit.

$ \displaystyle \lim_{x \to 0} x^2 = 0 $


&\text { Given } e>0, \text { we need } \delta>0 \text { such that if } 0<|x-0|<\delta, \text { then }\left|x^{2}-0\right|<\epsilon \Leftrightarrow x^{2}<\varepsilon \Leftrightarrow|x|<\sqrt{\varepsilon} . \text { Take } \delta=\sqrt{\epsilon}\\
&\text { Then } 0<|x-0|<\delta \Rightarrow\left|x^{2}-0\right|<\varepsilon . \text { Thus, } \lim _{x \rightarrow 0} x^{2}=0 \text { by the definition of a limit. }

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Video Transcript

So in this problem were asked to use the epsilon delta definition of limit to prove With the limit as X approaches zero of x squared. Kind of missy. Let's clean that up is zero. All right. So, first of all, what is this epsilon delta definition say? Well, it says that for all, absolutely greater than zero. There exists Delta greater than zero. So that if x minus A is less than delta, then over as it implies that f of X minus L is less than epsilon. So What a B0 L b zero and F of X B X squared. I'm not absolutely greater than zero. Be given. Okay? So then let's choose delta equal to the square root of epsilon. All right. So then X zero because a zero which is the absolute value of X less than delta, which is the square root of epsilon. And We have f of X zero. Right? Mm hmm. Instead of. And here, I could say that this implies it affects my zero. Because L was zero is an F of X is X squared. So, this means X squared absolute X squared. All right. Is less than the absolute value of square root of epsilon squared. Well, that's the step Salon, isn't it? Because that slows growth zero anyway. So, I don't take the square or not. Take. That's the value of it. And the square the square root that just means you get what's underneath the radical there. That's excellent. So, that means I have The limit as X goes to zero then by definition of f of X, which is x squared is zero, and there is our proof.