Proving a limit with the epsilon-delta definition $\lim\limits_{x \rightarrow 0} e^x = 1$

Question

I have to prove this limit with the epsilon-delta definition $$\lim\limits_{x \rightarrow 0} e^x = 1$$

I know I have to use $|x| < \delta$ and I have to get $|e^x-1|< \epsilon$

Any help?

Answer 1

Solution

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Preliminary considerations

Symbolically, we want to take the inequality $|e^x−1|<\epsilon$ and reach out the form $|x−0|<\delta$.

Thus

\begin{eqnarray*}|e^x - 1| < \epsilon & \\ -\epsilon < e^x - 1 < \epsilon & \qquad \textrm{(Definition of absolute value)} \\ 1-\epsilon < e^x < 1+ \epsilon & \qquad \textrm{(Add 1)} \\ \ln(1-\epsilon) < x < \ln(1+\epsilon). & \qquad \textrm{(Take natural logs)} \\ \end{eqnarray*}

Making the safe assumption that $\epsilon<1$ ensures the last inequality is valid (so that $\ln(1−\epsilon)$ is defined). We can then set $\delta$ to be the minimum of $|\ln(1−\epsilon)|$ and $\ln(1+\epsilon)$, as

$$\delta=\min\{|\ln(1−\epsilon)|,\ln(1+\epsilon)\}=\ln(1+\epsilon)$$

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Proof

\begin{align*} |x - 0|&<\delta\\ -\delta &< x < \delta & \textrm{(Definition of absolute value)}\\ -\ln(1+\epsilon) &< x < \ln(1+\epsilon) &\\ \ln(1-\epsilon) &< x < \ln(1+\epsilon). & \text{(since \(\ln(1-\epsilon) < -\ln(1+\epsilon)\))}\\ \end{align*}

The above line is true by our choice of $\delta$ and by the fact that since $|\ln(1−\epsilon)|>\ln(1+\epsilon)$ and $\ln(1−\epsilon)<0$, we know $\ln(1−\epsilon)<−\ln(1+\epsilon).$

\begin{align*}1-\epsilon &< e^x < 1+\epsilon & \textrm{(Exponentiate)}\\ -\epsilon &< e^x - 1 < \epsilon & \textrm{(Subtract 1)}\\ \end{align*}

In summary, given $\epsilon>0$, let $\delta=\ln(1+\epsilon)$. Then $|x−0|<\delta\Longrightarrow|e^x−1|<\epsilon$ as desired. So now, we have shown that $\displaystyle\lim_{x\rightarrow 0}e^x=1.$

Answer 2

Hint: See the third answer to this question which proves that $e^{x}-1 \ge x$ for every $x \in \mathbb{R}$.