Question

1. -/4 points prove the statement using the ε, δ definition of a limit. lim┬(x→0)⁡〖x^2 = 0〗 given ε > 0, we need δ select such that if 0 < |x - 0| < select, then |x^2 - 0| select. we have |x^2| < ε ⇔ x^2 < ε ⇔ |x| < √ε. choose δ = √ε. then 0 < |x - 0| < δ = |x^2 - 0| select. by the definition of a limit, lim┬(x→0)⁡〖x^2 = 0〗.

Explanation:

Step1: Recall limit - definition

The $\epsilon - \delta$ definition of $\lim_{x
ightarrow a}f(x)=L$ states that for every $\epsilon>0$, there exists a $\delta > 0$ such that if $0<|x - a|<\delta$, then $|f(x)-L|<\epsilon$. Here, $a = 0$, $f(x)=x^{2}$, and $L = 0$.

Step2: Set up the inequality

We want to find $\delta$ such that if $0<|x - 0|<\delta$, then $|x^{2}-0|<\epsilon$. Since $|x^{2}-0|=|x^{2}| = x^{2}$ (because $x^{2}\geq0$ for all real $x$), we have the inequality $x^{2}<\epsilon$.

Step3: Solve for $\delta$

Taking the square - root of both sides of $x^{2}<\epsilon$ (and considering only the positive square - root since $\delta>0$), we get $|x|<\sqrt{\epsilon}$. So, we choose $\delta=\sqrt{\epsilon}$. Then, if $0<|x - 0|=|x|<\delta=\sqrt{\epsilon}$, we have $|x^{2}-0|=x^{2}<\epsilon$.

Answer:

We have shown that for every $\epsilon>0$, choosing $\delta = \sqrt{\epsilon}$ satisfies the $\epsilon-\delta$ definition of the limit. So, $\lim_{x
ightarrow0}x^{2}=0$.