Calculating limits of function
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On practical classes at university we calculated some limits of functions of the following type:
$$lim_{xto 0}f(x)^{k(x)}$$
The key idea was that we know that $exists lim_{xto 0}f(x) = p in mathbb{R}$ (usually $p = e$). So we put this result in our initial expression:
$$lim_{xto 0}f(x)^{k(x)} = lim_{xto 0}p^{k(x)}$$
How can this step be proved formally? Probably, to prove this should be used variable substitution, but in such case I have ho ideas how to deal with $k(x)$.
upd: Ok, this cannot be proved in general, but what if $f(x) =(1+x)^{frac{1}{x}}$?
real-analysis limits
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up vote
1
down vote
favorite
On practical classes at university we calculated some limits of functions of the following type:
$$lim_{xto 0}f(x)^{k(x)}$$
The key idea was that we know that $exists lim_{xto 0}f(x) = p in mathbb{R}$ (usually $p = e$). So we put this result in our initial expression:
$$lim_{xto 0}f(x)^{k(x)} = lim_{xto 0}p^{k(x)}$$
How can this step be proved formally? Probably, to prove this should be used variable substitution, but in such case I have ho ideas how to deal with $k(x)$.
upd: Ok, this cannot be proved in general, but what if $f(x) =(1+x)^{frac{1}{x}}$?
real-analysis limits
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
On practical classes at university we calculated some limits of functions of the following type:
$$lim_{xto 0}f(x)^{k(x)}$$
The key idea was that we know that $exists lim_{xto 0}f(x) = p in mathbb{R}$ (usually $p = e$). So we put this result in our initial expression:
$$lim_{xto 0}f(x)^{k(x)} = lim_{xto 0}p^{k(x)}$$
How can this step be proved formally? Probably, to prove this should be used variable substitution, but in such case I have ho ideas how to deal with $k(x)$.
upd: Ok, this cannot be proved in general, but what if $f(x) =(1+x)^{frac{1}{x}}$?
real-analysis limits
On practical classes at university we calculated some limits of functions of the following type:
$$lim_{xto 0}f(x)^{k(x)}$$
The key idea was that we know that $exists lim_{xto 0}f(x) = p in mathbb{R}$ (usually $p = e$). So we put this result in our initial expression:
$$lim_{xto 0}f(x)^{k(x)} = lim_{xto 0}p^{k(x)}$$
How can this step be proved formally? Probably, to prove this should be used variable substitution, but in such case I have ho ideas how to deal with $k(x)$.
upd: Ok, this cannot be proved in general, but what if $f(x) =(1+x)^{frac{1}{x}}$?
real-analysis limits
real-analysis limits
edited Nov 26 at 18:23
asked Nov 26 at 16:01
Yan Kardziyaka
83
83
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This can't be proven formally, because it isn't true in general. Take $f(x) = 1 - x$, and $k(x) = frac{1}{x}$. Then $p = 1$, and so $lim_{xto 0}p^{k(x)} = 1$, but $lim_{xto 0}f(x)^{k(x)} = frac{1}{e}$.
thanks. Can you find counterexample for $f(x) = (1 + x)^frac{1}{x}$ ?
– Yan Kardziyaka
Nov 26 at 16:22
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
This can't be proven formally, because it isn't true in general. Take $f(x) = 1 - x$, and $k(x) = frac{1}{x}$. Then $p = 1$, and so $lim_{xto 0}p^{k(x)} = 1$, but $lim_{xto 0}f(x)^{k(x)} = frac{1}{e}$.
thanks. Can you find counterexample for $f(x) = (1 + x)^frac{1}{x}$ ?
– Yan Kardziyaka
Nov 26 at 16:22
add a comment |
up vote
2
down vote
accepted
This can't be proven formally, because it isn't true in general. Take $f(x) = 1 - x$, and $k(x) = frac{1}{x}$. Then $p = 1$, and so $lim_{xto 0}p^{k(x)} = 1$, but $lim_{xto 0}f(x)^{k(x)} = frac{1}{e}$.
thanks. Can you find counterexample for $f(x) = (1 + x)^frac{1}{x}$ ?
– Yan Kardziyaka
Nov 26 at 16:22
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
This can't be proven formally, because it isn't true in general. Take $f(x) = 1 - x$, and $k(x) = frac{1}{x}$. Then $p = 1$, and so $lim_{xto 0}p^{k(x)} = 1$, but $lim_{xto 0}f(x)^{k(x)} = frac{1}{e}$.
This can't be proven formally, because it isn't true in general. Take $f(x) = 1 - x$, and $k(x) = frac{1}{x}$. Then $p = 1$, and so $lim_{xto 0}p^{k(x)} = 1$, but $lim_{xto 0}f(x)^{k(x)} = frac{1}{e}$.
answered Nov 26 at 16:06
user3482749
2,086414
2,086414
thanks. Can you find counterexample for $f(x) = (1 + x)^frac{1}{x}$ ?
– Yan Kardziyaka
Nov 26 at 16:22
add a comment |
thanks. Can you find counterexample for $f(x) = (1 + x)^frac{1}{x}$ ?
– Yan Kardziyaka
Nov 26 at 16:22
thanks. Can you find counterexample for $f(x) = (1 + x)^frac{1}{x}$ ?
– Yan Kardziyaka
Nov 26 at 16:22
thanks. Can you find counterexample for $f(x) = (1 + x)^frac{1}{x}$ ?
– Yan Kardziyaka
Nov 26 at 16:22
add a comment |
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