Would this be considered a proper delta-epsilon proof?

Question

Is this a good delta-epsilon proof method? I think it's quite nice, but I'm not very advanced in math so I'm not sure if it's rigorous enough. $$\lim_{x\rightarrow 1}e^x=e$$ $$\begin{aligned} Proof. \text{ Let } c=1,L=e,f\left( x\right) =e^{x},f^{-1}\left( x\right) =\ln \left( x\right) ,f''\left( x\right) =e^{x}\end{aligned}$$ $$\begin{aligned} f''\left( x\right) >0\forall x\in \mathbb{R}\rightarrow 0< \delta \leq \left| f^{-1}\left( L+\varepsilon \right) -c\right|\end{aligned}$$ $$\begin{aligned} \left| f^{-1}\left( L+\varepsilon \right) -c\right| =\left| f^{-1}\left( \left( e\right) +\varepsilon \right) -\left( 1\right) \right|\end{aligned}$$ $$\begin{aligned} =\left| f^{-1}\left( e+\varepsilon \right) -1\right| =\left| \ln \left( e+\varepsilon \right) -1\right| =\ln \left(e+\varepsilon \right) -1\forall \varepsilon >0 \end{aligned}$$ $$\begin{aligned} \rightarrow 0 <\delta \leq \ln \left( e+\varepsilon \right) -1\forall \varepsilon >0\\ \therefore \forall \delta \in \mathbb{R} \exists \varepsilon >0\end{aligned}$$

Answer 1

It can be simpler than that. All you have to prove is that $\displaystyle \lim_{y \to 0}e^{y}=1$.

Take $\epsilon>0$ then you need to have that $-\epsilon<e^{y}-1<\,\epsilon$ which is $1-\epsilon<e^{y}<1+\epsilon$.

Take as $\delta$=ln$(1+\epsilon)$. Then $y<\delta$ implies $e^{y}<1+\epsilon$.

We also have that $(1-\epsilon^{2})<1$ hence ln$((1-\epsilon)(1+\epsilon))<0$

and ln$(1-\epsilon)$<$-$ln$(1+\epsilon)$.

Therefore $-$ln$(1+\epsilon)<y\,\,\,$ implies ln$(1-\epsilon)<y$ and hence

$1-\epsilon<\,e^{y}$. So taking $\delta$=ln$(1+\epsilon)$ we get :

$-\epsilon<e^{y}-1<\epsilon$ and hence $\displaystyle \lim_{y \to 0}e^{y}=1$. Setting $y=x-1$ we get $\displaystyle \lim_{x \to 1}e^{x-1}=1$

and $\displaystyle \lim_{x \to 1}e^{x}=e$

Answer 2

An alternative approach, that bypasses any use of logarithms, and allows you to directly analyze the behavior of $f(x) = e^x$, as $x$ approaches $(x = 1),$ both from below $(x = 1)$ and above $(x = 1)$ is as follows:

Use that the derivative is $f'(x) = e^x.$

What you have to show is that for any $\epsilon > 0$, you can find $\delta_1,\delta_2$, both positive, both in terms of $\epsilon$, such that:

• If $1 - \delta_1 < x < 1,$ then $|e^x - e| < \epsilon.$

• If $1 < x < 1 + \delta_2,$ then $|e^x - e| < \epsilon.$

Having shown both bullet points above, you would then satisfy the traditional $\epsilon,\delta$ definition of a limit by specifying that $\delta = \min(\delta_1,\delta_2).$

Throughout the remainder of this answer, it will be assumed that the artificial constraints of $0 < \delta_1, \delta_2 \leq \dfrac{1}{2}$ will be imposed.

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When $x = (1 - \delta_1)$, the rate of change of $f(x)$ is $e^{1 - \delta_1}$.

When $x = 1$, the rate of change of $f(x)$ is $e^1.$

Therefore, for $x$ in the interval $1 - \delta_1 < x < 1$, the rate of change of $f(x)$ is always positive, always increasing, and is strictly less than $e$. Therefore, it is impossible for any $x$ in that interval to have the distance of $|f(x) - f(1)|$ be greater than or equal to

$$\delta_1 \times e^1.$$

Therefore, if you choose $\delta_1$ so that $\delta_1 \times e^1 \leq \epsilon$, you will have fulfilled the requirement that

$$1 - \delta_1 < x < 1 \implies |f(x) - f(1)| < \epsilon.$$

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For $1 < x < 1 + \delta_2$, the analysis is very similar. The rate of change will always be strictly less than $e^{1 + \delta_2}.$

Therefore, it is impossible for $x$ in this interval to be such that

$$|f(x) - f(1)| \geq \delta_2 \times e^{1 + \delta_2}.$$

Keeping in mind the artificial constraint that $0 < \delta _2 \leq \dfrac{1}{2}$, you have that

$$\delta_2 \times e^{1 + \delta_2}$$

must be $~ \leq \delta_2 \times e^{1.5}.$

Therefore, all that you have to do, to complete the proof, is to choose $\delta_2$ so that

$$\delta_2 \times e^{1.5} = \epsilon.$$

Putting the various considerations of $\delta_1, \delta_2$ all together, the following specification for $\delta$ must work:

$$\delta = \min\left[\dfrac{1}{2}, \dfrac{\epsilon}{e^{1.5}}\right].$$