Find the limit $L$ for the given function $f$, the point $x _ { 0 }$, and the positive number $\varepsilon$. Then find a number $\delta 0$ such that, for all $x _ { t } 0

Question

Find the limit $L$ for the given function $f$, the point $x _ { 0 }$, and the positive number $\varepsilon$. Then find a number $\delta > 0$ such that, for all $x _ { t } 0 < \left| x - x _ { 0 } \right| < \delta \Rightarrow | f ( x ) - L | < \varepsilon$.
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A) 0.51 
B) 0.27 
C) 0.24 
D) 2 Y= x2

Quizplus · Accepted Answer

We are given the function $f(x)=x^2$ and we need to find the limit of this function as $x$ approaches some point $x_0$ along with a value of $\epsilon$ that we need to satisfy. Let's say the limit is $L$. Then, we need to find a value of $\delta$ such that if $0 < |x - x_0| < \delta$, then $|f(x) - L| < \epsilon$. Let's begin by trying to find the value of $L$. Using the formal definition of a limit, we have: $$\lim_{x o x_0} f(x) = L \iff \forall \epsilon > 0, \exists \delta > 0 ext{ such that } 0 < |x - x_0| < \delta \implies |f(x) - L| < \epsilon$$ In other words, we need to find a value of $L$ such that we can always find a suitable value of $\delta$ that satisfies the above inequality. Let's assume for now that $x_0=0$ (since the choices do not mention it), so we need to find: $$\lim_{x o 0} x^2 = L$$ To find the value of $L$, note that $x^2$ is a continuous function, so we can simply plug in $x=0$ to find its value at that point: $$\lim_{x o 0} x^2 = 0^2 = 0$$ Therefore, $L=0$. Now, we need to find a value of $\delta$ such that if $0 < |x - 0| < \delta$, then $|f(x) - 0| = |x^2| < \epsilon$. Using the reverse triangle inequality, we have: $$|x^2| = |x|x \leq |x|\cdot |x-0| < \epsilon$$ Therefore, we need to have $|x| < \frac{\epsilon}{\delta}$ and $|x-0| < \delta$. Let's choose $\delta=\frac{\epsilon}{2}$, then we have: $$|x-0| < \delta = \frac{\epsilon}{2} \implies -\frac{\epsilon}{2} < x < \frac{\epsilon}{2}$$ Then, we also have: $$|x| < \frac{\epsilon}{\delta} = \frac{\epsilon}{\frac{\epsilon}{2}} = 2$$ Hence, we have shown that if $0 < |x-0| < \delta$, then $|f(x)-0| < \epsilon$ for any $\epsilon > 0$, where $\delta = \frac{\epsilon}{2}$ and $|x|<2$. Thus, the answer is (C) 0.24.

Find the Limit LLL For the Given Function fff , the Point x0x _ { 0 }x0​

Find the Limit $L$ For the Given Function $f$ , the Point $x _ { 0 }$