# Limits of a function

1. May 19, 2015

### masterchiefo

1. The problem statement, all variables and given/known data
1. f(x) = (1+x)1/x
lim x->0+ (1+x)1/x

2. lim x->0+ 10+ln(x)

2. Relevant equations

3. The attempt at a solution
1. lim x->0+ (1+x)1/x

(1+0+)1/0+

I really dont understand how the final answer is e... thanks for helping.

= e1

2. lim x->0+ 10+ln(x)

ln(0+) = -infinite
10 - infinite = - infinite
I dont understand why ln(x) is - infinite is there any rule for that or thats something I have to know by heart ?

2. May 19, 2015

### Svein

1. If you let $u=\frac{1}{x}$, then $\lim_{x\rightarrow 0+} (1+x)^{\frac{1}{x}}$ transforms into $\lim_{u\rightarrow +\infty}(1+\frac{1}{u})^{u}$. Does this expression look familiar? (http://en.wikipedia.org/wiki/E_(mathematical_constant)).
2. Do the same transformation here ( $u=\frac{1}{x}$). The expression changes to $\lim_{u\rightarrow +\infty}(10+ln(\frac{1}{u}))=10+\lim_{u\rightarrow +\infty}ln(\frac{1}{u})=10-\lim_{u\rightarrow +\infty}ln(u)$.

3. May 19, 2015

### SammyS

Staff Emeritus
For item #2.

I don't know whether or not you want to consider it as " knowing by heart " , but if you're serious about math, physics, engineering, etc., you should be very familiar with the characteristics of logarithmic functions.

ln(x) is strictly increasing, it's domain is the interval $\displaystyle \ (0,\ \infty) \$, it's range is $\displaystyle \ (-\infty,\ \infty) \$.

That's enough to get $\displaystyle \ \lim_{x\to0^+} \ln(x)=-\infty \$.

I or someone else will cover item #1 in a separate post.

4. May 19, 2015

### nuuskur

$x = e^{\ln x}$. In our current problem, $x$ represents the expression your are taking the limit of.
In the process $x\to 0$ , the expression $\ln (1+x)$ is equivalent to $x$.

E: This is basic mathematical analysis and not about physics. Not being mean, just pointing it out.

Also, usually $x= e^{\ln |x|}$, however here we approach $0$ strictly from the right side hence the $x$ is positive regardless.

Last edited: May 19, 2015
5. May 19, 2015

### vela

Staff Emeritus
You should be able to reason it out. The relation $y = \ln x$ is equivalent to $x = e^y$. If you want to make x very very small, i.e. close to 0, what kind of value do you have to use for y?