Limits of a Function: Find e1 & -Infinite

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Homework Statement


1. f(x) = (1+x)1/x
lim x->0+ (1+x)1/x

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

Homework Equations

The Attempt at a Solution


1. lim x->0+ (1+x)1/x

(1+0+)1/0+

I really don't understand how the final answer is e... thanks for helping.

= e12. lim x->0+ 10+ln(x)

ln(0+) = -infinite
10 - infinite = - infinite
I don't understand why ln(x) is - infinite is there any rule for that or that's something I have to know by heart ?
 
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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).
 
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masterchiefo said:

Homework Statement


1. f(x) = (1+x)1/x
lim x->0+ (1+x)1/x

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

Homework Equations

The Attempt at a Solution


1. lim x->0+ (1+x)1/x

(1+0+)1/0+

I really don't understand how the final answer is e... thanks for helping.

= e12. lim x->0+ 10+ln(x)

ln(0+) = -infinite
10 - infinite = - infinite
I don't understand why ln(x) is - infinite is there any rule for that or that's something I have to know by heart ?
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.
 
##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.
 
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masterchiefo said:
I don't understand why ln(x) is -infinity. Is there any rule for that or is that something I have to know by heart?
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?
 
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