How do you prove that the exponential function is continuous at x=0 and how do you prove it is continuous for all real x?

faois3nh
2022-10-21
Answered

You can still ask an expert for help

faux0101d

Answered 2022-10-22
Author has **21** answers

Suppose that exp x is continuous at x=0. This means that

$$\underset{h\to 0}{lim}{e}^{h}=1$$

Now, consider any other x. Then

$${e}^{x}\cdot \underset{h\to 0}{lim}{e}^{h}={e}^{x}\phantom{\rule{0ex}{0ex}}\underset{h\to 0}{lim}{e}^{x}\cdot {e}^{h}={e}^{x}\phantom{\rule{0ex}{0ex}}\underset{h\to 0}{lim}{e}^{x+h}={e}^{x}$$

which says this is continuous at every other point. The same idea works in showing it is differentiable at every point as soon as it is differentiable at the origin. Now, it depends on how you define it to devise a proof of continuity and/or differentiability at the origin.

$$\underset{h\to 0}{lim}{e}^{h}=1$$

Now, consider any other x. Then

$${e}^{x}\cdot \underset{h\to 0}{lim}{e}^{h}={e}^{x}\phantom{\rule{0ex}{0ex}}\underset{h\to 0}{lim}{e}^{x}\cdot {e}^{h}={e}^{x}\phantom{\rule{0ex}{0ex}}\underset{h\to 0}{lim}{e}^{x+h}={e}^{x}$$

which says this is continuous at every other point. The same idea works in showing it is differentiable at every point as soon as it is differentiable at the origin. Now, it depends on how you define it to devise a proof of continuity and/or differentiability at the origin.

Cory Russell

Answered 2022-10-23
Author has **4** answers

If you know that the exponential function is differntiable on R,

$$(\mathrm{exp}{)}^{\prime}(x)=\mathrm{exp}(x),\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in \mathbb{R},$$

then it's continuous on R as well.

$$(\mathrm{exp}{)}^{\prime}(x)=\mathrm{exp}(x),\phantom{\rule{1em}{0ex}}\mathrm{\forall}x\in \mathbb{R},$$

then it's continuous on R as well.

asked 2022-10-07

Prove for any real number k, prove that the exponential function ${e}^{z}$ is a bijection (z is a complex number) from the strip $k<Im(z)\le k+2\pi $ to the complex plane minus the point 0, $\mathbb{C}-\{0\}$

asked 2022-06-24

asked 2022-10-11

Why is the expectation of an exponential function:

$$\mathbb{E}[\mathrm{exp}(Ax)]=\mathrm{exp}((1/2){A}^{2})\phantom{\rule{thinmathspace}{0ex}}?$$

$$\mathbb{E}[\mathrm{exp}(Ax)]=\mathrm{exp}((1/2){A}^{2})\phantom{\rule{thinmathspace}{0ex}}?$$

asked 2022-11-04

Given the vector space, $C(-\mathrm{\infty},\mathrm{\infty})$ as the set of all continuous functions that are always continuous, is the set of all exponential functions, $U=\{{a}^{x}\mid a\ge 1\}$, a subspace of the given vector space?**As far as I'm aware, proving a subspace of a given vector space only requires you to prove closure under addition and scalar multiplication, but I'm kind of at a loss as to how to do this with exponential functions (I'm sure it's way simpler than I'm making it).****My argument so far is that the set U is a subset of the set of all differentiable functions, which itself is a subset of $C(-\mathrm{\infty},\mathrm{\infty})$, but I doubt that argument would hold up on my test, given how we've tested for subspaces in class (with closure).**

asked 2022-10-17

The exponential function is the unique function satisfying

f′=f and f(0)=1

however, unless I've made a mistake, we have

$$\frac{\mathrm{\partial}}{\mathrm{\partial}x}(ax{)}^{x}=x(ax{)}^{x-1}a=ax(ax{)}^{x-1}=(ax{)}^{x}$$

and

$$(a0{)}^{0}={0}^{0}=1$$

so I feel like I must be missing something special about ${e}^{x}$. Any pointers would be greatly appreciated.

f′=f and f(0)=1

however, unless I've made a mistake, we have

$$\frac{\mathrm{\partial}}{\mathrm{\partial}x}(ax{)}^{x}=x(ax{)}^{x-1}a=ax(ax{)}^{x-1}=(ax{)}^{x}$$

and

$$(a0{)}^{0}={0}^{0}=1$$

so I feel like I must be missing something special about ${e}^{x}$. Any pointers would be greatly appreciated.

asked 2022-11-13

The hyperbolic functions can be expressed using the exponential function. However how are these related to "hyperbolas"?

asked 2022-09-06

I have used the quotient rule and chain rule to solve for ${0.6}^{x}$ but its still wrong.

$$P(x)=\frac{1864}{1+49\times (0.6{)}^{x}}$$

$$P(x)=\frac{1864}{1+49\times (0.6{)}^{x}}$$