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Question: $f(x) = e^{x}$ at $x = 0$:

$\forall\epsilon>0, \exists\delta>0 \text{ s.t. } |x - 0|<\delta \implies |f(x) - f(0)|<\epsilon$

$$\begin{align}|f(x) - f(0)| &= |e^{x} - 1| \\ &\le e^{x} + 1 \\ &< e^{\delta} + 1 \end{align}$$

Take $\epsilon = e^{\delta} + 1$ so $\delta = \ln{(\epsilon -1)}$ but this doesn't give a $\delta > 0$ if $\epsilon \le 2$, what have I done wrong?

P.s. I want to do this without using the fact that $\displaystyle e^{x} = \sum^{\infty}_{k=0} \frac{x^{k}}{k!}$ and I don't want to use limits either.

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  • $\begingroup$ Well, for starters, $e^x + 1 > 1$ always, so the problem is going to be in the jump from $|e^x - 1|$ to $e^x + 1$. $\endgroup$ – Neal Apr 30 '14 at 23:52
  • $\begingroup$ Why is this a problem? $|e^{x}-1| \le e^{x}+1 \quad \forall x$ $\endgroup$ – user2850514 Apr 30 '14 at 23:56
  • $\begingroup$ @user2850514 The problem is that it tells you nothing; this was remarked on in your previous question. $\endgroup$ – user61527 Apr 30 '14 at 23:57
  • $\begingroup$ Yes but nobody actually told me what else to do which is why I made a new post. I stated I don't want any summation or limits. $\endgroup$ – user2850514 Apr 30 '14 at 23:59
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Instead of using $|e^x - 1| < e^x + 1$ (which is practically trivial), note that you want the inequality $|e^x - 1| < \epsilon$. Rephrasing this, $$ 1-\epsilon < e^x < 1+\epsilon.$$ Can you see how to take it from here?

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  • $\begingroup$ So we have $\ln{(1 - \epsilon) < x < \ln{(1 + \epsilon)}}$. Take $\delta = \ln{(1 + \epsilon)}$, is this correct? $\endgroup$ – user2850514 Apr 30 '14 at 23:58
  • $\begingroup$ @user2850514 You should be able to check it yourself ;) (I would take into account that $|\ln(1-\epsilon)|\neq |\ln(1+\epsilon)|$ in choosing $\delta$.) $\endgroup$ – Neal May 1 '14 at 0:08
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    $\begingroup$ Dear @user2850514, yes that is almost correct (except that $\delta$ should also be less than $\left|\ln (1-\epsilon)\right|$), assuming $\epsilon<1$ (which you can do without loss of generality) and using the monotonicity of the $\ln$ function. I really like this answer (+1) but it does, in a sense, sweep under the rug the use of some non-trivial properties of the exponential function (for example, how to prove that $\ln$ is monotonically increasing?). My answer below explicitly uses the differentiability of the exponential function at $x=0$ but is more indirect - please have a look at it too. $\endgroup$ – Amitesh Datta May 1 '14 at 0:09
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    $\begingroup$ So $$\delta = \begin{cases} \ln{(1-\epsilon)} & \epsilon < 1 \\ \ln{(1 + \epsilon)} & \epsilon \ge 1 \end{cases}$$ is correct? $|x|<\ln{(1 + \epsilon)} \implies |e^{x}| < 1+\epsilon \implies |e^{x}-1|<\epsilon$ $\endgroup$ – user2850514 May 1 '14 at 0:17
  • $\begingroup$ and $|x|<\ln{(1 - \epsilon)} \implies |e^{x}| < 1-\epsilon \implies |e^{x}-1|<\epsilon$ $\endgroup$ – user2850514 May 1 '14 at 0:23
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You have to somehow use the fact that you are dealing with $f(x)=e^x$ and not some arbitrary function. In other words, you have to use something about the exponential function that you know, that implies continuity at $x=0$. Thus, you have to either use the definition of the exponential function (near $x=0$), or some property of the exponential function near $x=0$ that follows from the definition.

Neal has already suggested an approach in his excellent answer. Alternatively, if you are willing to accept that $f(x)=e^x$ is differentiable at $x=0$, then continuity follows from differentiability as follows:

$$\lim_{x\to 0} \frac{e^x-1}{x}$$

exists, because we know that $f(x)=e^x$ is differentiable at $x=0$. Since $\lim_{x\to 0} x=0$, the product rule for limits implies

$$\lim_{x\to 0} \frac{e^x-1}{x}\cdot x = 0,$$ i.e.,

$$\lim_{x\to 0} (e^x - 1)=0.$$

Now, use the difference rule for limits to conclude that $\lim_{x\to 0} e^x = 1$. (Notice in this approach, you didn't even need to know the value of the derivative $f'(0)$ - just knowing that the limit defining the derivative existed was enough.)

You might argue that proving differentiability is harder than proving continuity, and you would certainly be correct. However, I just wanted to give you an alternative perspective.

Exercise: Let $f:\mathbb{R}\to \mathbb{R}$ be a function that is differentiable at some $a\in \mathbb{R}$. Prove that $f$ is continuous at $a$. (Hint: mimic the proof in the above case of $f(x)=e^x$.)

Hope this helps!

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  • $\begingroup$ I said in the question I don't want a limit proof or using the summation, thanks for the effort though. $\endgroup$ – user2850514 May 1 '14 at 0:07
  • $\begingroup$ Thanks, @Michael, for improving the presentation of my answer through your edit! $\endgroup$ – Amitesh Datta May 1 '14 at 0:30
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The slope of $y=e^x$ at $x=0$ is $1$, and when $x$ is near $0$ the slope is near $1$. You have to get a certain distance from $x=0$ before the slope gets anywhere near $2$. Sufficiently near $x=0$, the slope is $\le 2$.

So let $\delta = \varepsilon/2$ for sufficiently small $\varepsilon$. And we only care about "sufficiently small" values of $\varepsilon$.

If $x$ is between $0-\delta$ and $0+\delta$ then $|e^x-x^0|$ is less than $2$ times $|x-0|$, so less than $2\delta=\varepsilon$.

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Assume $\lim_{n \to \infty} \sqrt[n]{e}=1$. If $\epsilon \gt 0$ then there is an $N_\epsilon \in \mathbb{N}$ such that $|e^\frac{1}{n}-1|<\epsilon$ if $n \ge N_\epsilon$. So $|e^x-1|<\epsilon$ if $x \lt \delta:= \frac{1}{N_\epsilon}$ because $e^x$ is increasing if $x \in \mathbb{R^+}$.

Why do we know that $\lim_{n \to \infty} \sqrt[n]{e}=1$?

From Bernoulli's inequality $(1+r)^n>=1+rn$ we get

$$1 \lt e^\frac {1}{n}=\left( 1+\left(e-1\right) \right)^\frac{1}{n}\le1+\frac{e-1}{n}$$

Why do we know that $e^x$ is increasing?

We know that $e^x$ is increasing if $x \in \mathbb{Q^+}$ therefore its continuation to $\mathbb{R^+}$ is increasing to

This proof works for every $e>1$

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