MHB How can I show with the definition that f is continuous?

[evinda]

Hello! (Smile)
I am given this exercise:
$$f(x)=\left\{\begin{matrix}
\frac{e^x-1}{x} &, x \neq 0 \\
1& ,x=0
\end{matrix}\right. , x \in [0,1]$$

Show that $f$ is integrable in $[0,1]$,knowing that if $f:[a,b] \to \mathbb{R}$, $f$ continuous,then $f$ is integrable in $[a,b]$.

So,I have to show that $f$ is continuous at the whole interval $[0,1]$.

It is: $\lim_{x \to 0} \frac{e^x-1}{x}=\lim_{x \to 0} e^x=1=f(0)$
So, $f$ is continuous at $0$

But how can I show it,using the definition of the continuity?

Let $\epsilon>0$.
We want to show that $\exists \delta>0$ such that $\forall x \in [0,1]$ with $|x-0|<\delta \Rightarrow |f(x)-f(0)| < \epsilon \Rightarrow |f(x)|< \epsilon$

How can I continue?

 

[Prove It]

Well first of all, if $\displaystyle \begin{align*} \left| f(x) - f(0) \right| < \epsilon \end{align*}$, then $\displaystyle \begin{align*} \left| f(x) - 1 \right| < \epsilon \end{align*}$, not $\displaystyle \begin{align*} \left| f(x) \right| \end{align*}$.

 

[chisigma]

Hello! (Smile)
I am given this exercise:
$$f(x)=\left\{\begin{matrix}
\frac{e^x-1}{x} &, x \neq 0 \\
1& ,x=0
\end{matrix}\right. , x \in [0,1]$$

Show that $f$ is integrable in $[0,1]$,knowing that if $f:[a,b] \to \mathbb{R}$, $f$ continuous,then $f$ is integrable in $[a,b]$.

So,I have to show that $f$ is continuous at the whole interval $[0,1]$.

It is: $\lim_{x \to 0} \frac{e^x-1}{x}=\lim_{x \to 0} e^x=1=f(0)$
So, $f$ is continuous at $0$

But how can I show it,using the definition of the continuity?

Let $\epsilon>0$.
We want to show that $\exists \delta>0$ such that $\forall x \in [0,1]$ with $|x-0|<\delta \Rightarrow |f(x)-f(0)| < \epsilon \Rightarrow |f(x)|< \epsilon$

How can I continue?

The limit...

$\displaystyle \lim_{x \rightarrow 0} \frac {e^{x}-1}{x}\ (1)$

... is demonstrated without using L'Hopital rule starting from the basic definition...$\displaystyle \lim_{\xi \rightarrow \infty} (1 + \frac{1}{\xi})^{\ \xi} = e\ (2)$

Setting $\displaystyle e^{x} = 1 + \frac{1}{\xi}$ we obtain...

$\displaystyle \lim_{x \rightarrow 0} \frac{e^{x} - 1}{x} = \lim_{\xi \rightarrow \infty} \frac{1}{\xi\ \ln (1 + \frac{1}{\xi})} = \lim_{\xi \rightarrow \infty} \frac{1}{ \ln (1 + \frac{1}{\xi})^ {\xi}} = \frac{1}{\ln e} =1\ (3)$

Kind regards$\chi$ $\sigma$

 

[Prove It]

Since $\displaystyle \begin{align*} e^x = \sum_{n = 0}^{\infty} \frac{x^n}{n!} \end{align*}$, that means

$\displaystyle \begin{align*} e^x &= 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots \\ e^x - 1 &= x + \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots \\ \frac{e^x - 1}{x} &= 1 + \frac{x}{2} + \frac{x^2}{3!} + \frac{x^3}{4!} +\dots \end{align*}$

So if $\displaystyle \begin{align*} \left| f(x) - f(0) \right| < \epsilon \end{align*}$ that means

$\displaystyle \begin{align*} \left| 1 + \frac{x}{2} + \frac{x^2}{3!} + \frac{x^3}{4!} + \dots - 1 \right| &< \epsilon \\ \left| \frac{x}{2} + \frac{x^2}{3!} + \frac{x^3}{4!} + \dots \right| &< \epsilon \\ \left| x \right| \left| \frac{1}{2} + \frac{x}{3!} + \frac{x^2}{4!} + \dots \right| &< \epsilon \end{align*}$

Now if we restrict $\displaystyle \begin{align*} |x| < 1 \end{align*}$, that means

$\displaystyle \begin{align*} \left| \frac{1}{2} + \frac{x}{3!} + \frac{x^2}{4!} + \dots \right| &< \left| \frac{1}{2} + \frac{1}{3!} + \frac{1}{4!} + \dots \right| \\ &= \left| 1 + \frac{1}{2} + \frac{1}{3!} + \frac{1}{4!} + \dots - 1 \right| \\ &= \left| e - 1 \right| \end{align*}$

So that means you can set $\displaystyle \begin{align*} \delta = \min \left\{ 1, \frac{\epsilon}{e - 1} \right\} \end{align*}$ and you can start the proof :)