# Limits? changing and transformation

1. Nov 11, 2012

### harjyot

okay, so I'm at the most elementary stage of learning limits and there are things which leave me baffled at times, namely two.
1. lim (x -> a) f(x) = lim (x+k -> a+k) f(x)

how? the physical reason behind this?

2. the theorem to evaluate limits of the form --- 1^infinity
if f(x)=g(x)=0 (lim: x->a)
that is,
lim (x -> a) [1+f(x)]^{1/g(x)} =
e^{lim x -> a. f(x)/g(x))

2. Nov 11, 2012

### LeonhardEuler

For 1, I don't know how technically you learned the definition of the limit, but you know the idea of the limit is that $\lim_{x\rightarrow a}f(x)=L$ means that f(x) is very close to L as long as x is close enough to a. "x is close enough to a" is literally the exact same condition as "x+k is close enough to a+k".

For 2, you know that one definition of e is
$$e=\lim_{x\rightarrow 0}(1+x)^{1/x}$$
Well,
$$(1+f(x))^{1/g(x)} = (1+f(x))^{\frac{f(x)}{f(x)g(x)}}=[(1+f(x))^{1/f(x)}]^{\frac{f(x)}{g(x)}}$$
If you call f(x) "t", you will see that as t->0, you get the limit to be e inside. You have to do a bit more proving to show that everything goes through as you would expect, but at least you see why it makes sens to expect that result.

3. Nov 12, 2012

### harjyot

yes, for 1 I get that x is close to a, so x+k is close to a+k, but the doubt I have is how the result Is same?

4. Nov 12, 2012

### MarneMath

One way to see the first property is let y = x + k, and thus f(x) = f(y-k). From there it is pretty straight forward. I'm a bit to lazy to find a formal proof, but I imagine the argument would follow that line. To be honest, I never came across that property, so I'm not entirely sure if it's always true, but it seems right.

5. Nov 12, 2012

### HallsofIvy

Staff Emeritus
$\lim_{x\to a}f(x)= L$ means that "'Given any $\epsilon> 0$, there exist $\delta> 0$ such that if $|x- a|< \delta$, then $|f(x)- L|< \epsilon$".

And it should be easy to see that we can replace x by x+ k and a by a+ k to get exactly the same result: |(x+k)- (a+ k)|= |x- a|.

Be careful, this does NOT say that the limit "at x= a+ k" is the same as the limit "at x= a". We are still dealing with |f(x)- L|, NOT with |f(x+k)- L|.

6. Nov 13, 2012

### harjyot

thank you! now it's clear. we just express a expression in some other way by changing the limit but the meaning and result remains the same