Let $f:mathbb{Q}to mathbb{R}$, $f(p/q)=a^{p/q}$ such that $a>1$. Prove that $lim_{xto 0}f(x)=1$
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I need to solve that problem. I tried proving instead $lim_{xto infty}a^{1/x}=1$. But the complete proble say:
Let $f:mathbb{Q}to mathbb{R}$, with $f(p/q)=a^{p/q}$ such that a>1. Prove that $lim_{xto 0}f(x)=1$ and conclude:
- For all $binmathbb{R}$, exists $lim_{xto b} f(x)$.
- If $bin mathbb{Q}$ then $lim_{xto b} f(x)=f(b)$.
- $a^x cdot a^y =a^{x+y}$
- If $x<y$ then $a^x < a^y$.
I use the fourth point but the problem tells me that i need to conclude it from the first statement. I really don't know how to proceed, may something like $varepsilon -delta$.
I'll apreciate your help.
calculus limits limits-without-lhopital
asked Jan 8 at 7:52
JMartinJMartin
62
$endgroup$
add a comment |
$begingroup$
I need to solve that problem. I tried proving instead $lim_{xto infty}a^{1/x}=1$. But the complete proble say:
Let $f:mathbb{Q}to mathbb{R}$, with $f(p/q)=a^{p/q}$ such that a>1. Prove that $lim_{xto 0}f(x)=1$ and conclude:
- For all $binmathbb{R}$, exists $lim_{xto b} f(x)$.
- If $bin mathbb{Q}$ then $lim_{xto b} f(x)=f(b)$.
- $a^x cdot a^y =a^{x+y}$
- If $x<y$ then $a^x < a^y$.
I use the fourth point but the problem tells me that i need to conclude it from the first statement. I really don't know how to proceed, may something like $varepsilon -delta$.
I'll apreciate your help.
calculus limits limits-without-lhopital
asked Jan 8 at 7:52
JMartinJMartin
62
$endgroup$
1
$begingroup$
Using my powers of mind reading to understand what the author meant, I guess that for some reason, between point 2. and point 3., $x$ and $y$ suddenly became real. Or something. Showing that 3. and 4. holds for rational $x, y$ is almost trivial. Showing that they hold for real $x$ and $y$ (with $f$ and limits being used to define $a$ raised to a real number, and already knowing that they hold for rational exponents) would be a more suitable exercise at this stage.
$endgroup$
– Arthur
Jan 8 at 8:18
add a comment |
$begingroup$
I need to solve that problem. I tried proving instead $lim_{xto infty}a^{1/x}=1$. But the complete proble say:
Let $f:mathbb{Q}to mathbb{R}$, with $f(p/q)=a^{p/q}$ such that a>1. Prove that $lim_{xto 0}f(x)=1$ and conclude:
- For all $binmathbb{R}$, exists $lim_{xto b} f(x)$.
- If $bin mathbb{Q}$ then $lim_{xto b} f(x)=f(b)$.
- $a^x cdot a^y =a^{x+y}$
- If $x<y$ then $a^x < a^y$.
I use the fourth point but the problem tells me that i need to conclude it from the first statement. I really don't know how to proceed, may something like $varepsilon -delta$.
I'll apreciate your help.
calculus limits limits-without-lhopital
asked Jan 8 at 7:52
JMartinJMartin
62
$endgroup$
I need to solve that problem. I tried proving instead $lim_{xto infty}a^{1/x}=1$. But the complete proble say:
Let $f:mathbb{Q}to mathbb{R}$, with $f(p/q)=a^{p/q}$ such that a>1. Prove that $lim_{xto 0}f(x)=1$ and conclude:
- For all $binmathbb{R}$, exists $lim_{xto b} f(x)$.
- If $bin mathbb{Q}$ then $lim_{xto b} f(x)=f(b)$.
- $a^x cdot a^y =a^{x+y}$
- If $x<y$ then $a^x < a^y$.
I use the fourth point but the problem tells me that i need to conclude it from the first statement. I really don't know how to proceed, may something like $varepsilon -delta$.
I'll apreciate your help.
calculus limits limits-without-lhopital
calculus limits limits-without-lhopital
asked Jan 8 at 7:52
JMartinJMartin
62
asked Jan 8 at 7:52
JMartinJMartin
62
edited Jan 8 at 8:06
JMartin
asked Jan 8 at 7:52
JMartinJMartin
62
asked Jan 8 at 7:52
JMartinJMartin
62
asked Jan 8 at 7:52
JMartinJMartin
62
62
1
$begingroup$
Using my powers of mind reading to understand what the author meant, I guess that for some reason, between point 2. and point 3., $x$ and $y$ suddenly became real. Or something. Showing that 3. and 4. holds for rational $x, y$ is almost trivial. Showing that they hold for real $x$ and $y$ (with $f$ and limits being used to define $a$ raised to a real number, and already knowing that they hold for rational exponents) would be a more suitable exercise at this stage.
$endgroup$
– Arthur
Jan 8 at 8:18
add a comment |
1
$begingroup$
Using my powers of mind reading to understand what the author meant, I guess that for some reason, between point 2. and point 3., $x$ and $y$ suddenly became real. Or something. Showing that 3. and 4. holds for rational $x, y$ is almost trivial. Showing that they hold for real $x$ and $y$ (with $f$ and limits being used to define $a$ raised to a real number, and already knowing that they hold for rational exponents) would be a more suitable exercise at this stage.
$endgroup$
– Arthur
Jan 8 at 8:18
1
1
$begingroup$
Using my powers of mind reading to understand what the author meant, I guess that for some reason, between point 2. and point 3., $x$ and $y$ suddenly became real. Or something. Showing that 3. and 4. holds for rational $x, y$ is almost trivial. Showing that they hold for real $x$ and $y$ (with $f$ and limits being used to define $a$ raised to a real number, and already knowing that they hold for rational exponents) would be a more suitable exercise at this stage.
$endgroup$
– Arthur
Jan 8 at 8:18
$begingroup$
Using my powers of mind reading to understand what the author meant, I guess that for some reason, between point 2. and point 3., $x$ and $y$ suddenly became real. Or something. Showing that 3. and 4. holds for rational $x, y$ is almost trivial. Showing that they hold for real $x$ and $y$ (with $f$ and limits being used to define $a$ raised to a real number, and already knowing that they hold for rational exponents) would be a more suitable exercise at this stage.
$endgroup$
– Arthur
Jan 8 at 8:18
add a comment |
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$begingroup$
Using my powers of mind reading to understand what the author meant, I guess that for some reason, between point 2. and point 3., $x$ and $y$ suddenly became real. Or something. Showing that 3. and 4. holds for rational $x, y$ is almost trivial. Showing that they hold for real $x$ and $y$ (with $f$ and limits being used to define $a$ raised to a real number, and already knowing that they hold for rational exponents) would be a more suitable exercise at this stage.
$endgroup$
– Arthur
Jan 8 at 8:18