# Upper bound on polynomial similar to the limit that defines the exponential function

I don't understand how to show that the following is true:

$$(1+\frac{\epsilon}{5})^t \leq e^{\frac{t\epsilon}{5}}$$

I know that the exponential function can be defined as:

$$\lim_{n\to \infty} (1+\frac{1}{n})^n = e$$ and then by substitution $$n \to nx$$ we can obtain that $$\lim_{n\to \infty} (1+\frac{x}{n})^n = \lim_{n\to \infty} (1+\frac{1}{n})^{nx} = e^x$$

Moreover, we also know that the first limit is non-decreasing, thus approaching the value of $$e$$ from below (thus it should be simple to show the upper bound). My problem is that $$\frac{\epsilon}{5}$$ does not depend on $$t$$, so I am not really sure what to do.

Of course, we can do a substitution with $$x$$ and say $$\frac{t\epsilon}{5} = x$$ and obtain:

$$(1 + \frac{t\epsilon}{5} \frac{1}{t})^t = e^{\frac{t\epsilon}{5}}$$

but it seems a little bit strange, as we don't change the exponent, nor we have another $$x$$ on the left hand side of the substitution (i.e. before we had $$x \rightarrow nx$$, here the $$x$$ is not appearing on one side).

Is this correct? How can I know that my substitusions are correct? Is there a list of rule / tutorial / book to study? Thanks.

Write $$\left(1+\frac\epsilon 5\right)^t=\left(1+\frac\epsilon 5\right)^{\frac 5\epsilon \frac\epsilon 5t}=\left[\left(1+\frac\epsilon 5\right)^{\frac 5\epsilon}\right]^{\frac{\epsilon t}5}\le e^{\frac{\epsilon t}5}$$

• I don't understand. Where is the limit? What goes to infinity here?
– asdf
Jul 7 at 18:22
• There is no limit. You just need to show $(1+\epsilon/5)^{5/\epsilon}\le e$ Jul 7 at 18:24
• See math.stackexchange.com/questions/167843/… for example Jul 7 at 18:35

If $$t$$ is an integer, use binomial expansion.

$$(1+\frac{x}{n})^n=\sum\limits_{k=0}^n \binom{n}{k}(\frac{x}{n})^k$$

$$e^x=\sum\limits_{k=0}^\infty \frac{1}{k!}x^k$$.

$$\binom{n}{k}\frac{1}{n^k}=\frac{n(n-1)(...(n-k+1)}{n^k}{\frac{1}{k!}}\le \frac{1}{k!}$$

Net result $$(1+\frac{x}{n})^n\le e^x$$.