Q. 2 - Calculus

Question

Explain why it makes intuitive sense that $\underset{x\to c}{lim} x^2=c^2$ for any real number c. Then use a delta–epsilon argument to prove it. (Hint: You will need to assume that δ ≤ 1 )

Step-by-Step Solution

Answer

The explanation has been made.

Step 1. Given information

We have to explain why it makes intuitive sense that $\underset{x\to c}{lim} x^2=c^2$ for any real number c. Then use a delta–epsilon argument to prove it. (Hint: You will need to assume that δ ≤ 1 )

Step 2. Explanation

Substitute c for x in the limit,

$\underset{x\to c}{lim} x^2=c^2$

For the limit statement $\underset{x\to c}{lim} f^{\mathrm{\prime }}\left(x\right)=L$, the delta-epsilon statement is

For all $\in >0$, there exist $\delta >0$ such that whenever

$x\in (c-\delta ,c)\cup (c,c+\delta )\text{ guarantees }f\left(x\right)\in (L-\in ,L+\in )$

For every x satisfying $0<|x-c|<\delta$, every f(x) satisfies $\left|f\right(x)-L|<ϵ$

$f\left(x\right)=x^2\text{ and }L=c^2$

Choose $\delta =\frac{ϵ}{2c+1}$

$\begin{array}{c}\left|f\right(x)-L|=\left|x^2-c^2\right|\\ =|x+c||x-c|\\ <\delta |x+c|\end{array}$

Now, for $0<|x-c|<\delta \text{ and }\delta <1\text{ substitute }x=1+c$

$\begin{array}{c}\left|f\right(x)-L|<\delta |x+c|\\ =\delta |1+c2|\\ =\frac{ϵ}{|1+2c|}|1+2c|\\ <\in \end{array}$