Mixing
Prove that $f'(0)=5$
$begingroup$
Let the function $h:mathbb R rightarrow mathbb R$ be bounded. Define $f:mathbb R rightarrow mathbb R$ by $f(x)=2+5x+x^2h(x)$. Prove that $f(0)=2$ and $f'(0)=5$.
I think$f(0)=2$ is pretty obvious. Next $$f'(0)= lim_{x to 0} frac{f(x)-f(0)}{x-0}=lim_{x to 0} frac{2+5x+x^2h(x)-2}{x}=lim_{x to 0}frac{5+2xh(x)+x^2h'(x)}{1}=5,$$ right? In that case why do we need h to be bounded?
real-analysis
asked Jan 27 at 4:28
dxdydzdxdydz
42110
$endgroup$
add a comment |
$begingroup$
Let the function $h:mathbb R rightarrow mathbb R$ be bounded. Define $f:mathbb R rightarrow mathbb R$ by $f(x)=2+5x+x^2h(x)$. Prove that $f(0)=2$ and $f'(0)=5$.
I think$f(0)=2$ is pretty obvious. Next $$f'(0)= lim_{x to 0} frac{f(x)-f(0)}{x-0}=lim_{x to 0} frac{2+5x+x^2h(x)-2}{x}=lim_{x to 0}frac{5+2xh(x)+x^2h'(x)}{1}=5,$$ right? In that case why do we need h to be bounded?
real-analysis
asked Jan 27 at 4:28
dxdydzdxdydz
42110
$endgroup$
4
$begingroup$
Try $h(x) = 1/x^2$ for $x ne 0$,$0$ for $x=0$ for a counter example.
$endgroup$
– D.B.
Jan 27 at 4:32
$begingroup$
Why the $2xh+x^2h$ in last equality ? Where does that comes from all of a sudden ?
$endgroup$
– zwim
Jan 27 at 5:07
$begingroup$
@zwim L'hopitale's rule, there was a typo, edited
$endgroup$
– dxdydz
Jan 27 at 18:59
1
$begingroup$
You are not allowed to use L'Hospital's Rule because you don't know anything about $h$ apart from it being bounded. The fraction under limit simplifies to $5+xh(x)$ and this tends to $5$ as $xh(x) to 0$ (and this is where we need $h$ to be bounded).
$endgroup$
– Paramanand Singh
Jan 27 at 19:25
add a comment |
$begingroup$
Let the function $h:mathbb R rightarrow mathbb R$ be bounded. Define $f:mathbb R rightarrow mathbb R$ by $f(x)=2+5x+x^2h(x)$. Prove that $f(0)=2$ and $f'(0)=5$.
I think$f(0)=2$ is pretty obvious. Next $$f'(0)= lim_{x to 0} frac{f(x)-f(0)}{x-0}=lim_{x to 0} frac{2+5x+x^2h(x)-2}{x}=lim_{x to 0}frac{5+2xh(x)+x^2h'(x)}{1}=5,$$ right? In that case why do we need h to be bounded?
real-analysis
asked Jan 27 at 4:28
dxdydzdxdydz
42110
$endgroup$
Let the function $h:mathbb R rightarrow mathbb R$ be bounded. Define $f:mathbb R rightarrow mathbb R$ by $f(x)=2+5x+x^2h(x)$. Prove that $f(0)=2$ and $f'(0)=5$.
I think$f(0)=2$ is pretty obvious. Next $$f'(0)= lim_{x to 0} frac{f(x)-f(0)}{x-0}=lim_{x to 0} frac{2+5x+x^2h(x)-2}{x}=lim_{x to 0}frac{5+2xh(x)+x^2h'(x)}{1}=5,$$ right? In that case why do we need h to be bounded?
real-analysis
real-analysis
asked Jan 27 at 4:28
dxdydzdxdydz
42110
asked Jan 27 at 4:28
dxdydzdxdydz
42110
edited Jan 27 at 18:58
dxdydz
asked Jan 27 at 4:28
dxdydzdxdydz
42110
asked Jan 27 at 4:28
dxdydzdxdydz
42110
asked Jan 27 at 4:28
dxdydzdxdydz
42110
42110
4
$begingroup$
Try $h(x) = 1/x^2$ for $x ne 0$,$0$ for $x=0$ for a counter example.
$endgroup$
– D.B.
Jan 27 at 4:32
$begingroup$
Why the $2xh+x^2h$ in last equality ? Where does that comes from all of a sudden ?
$endgroup$
– zwim
Jan 27 at 5:07
$begingroup$
@zwim L'hopitale's rule, there was a typo, edited
$endgroup$
– dxdydz
Jan 27 at 18:59
1
$begingroup$
You are not allowed to use L'Hospital's Rule because you don't know anything about $h$ apart from it being bounded. The fraction under limit simplifies to $5+xh(x)$ and this tends to $5$ as $xh(x) to 0$ (and this is where we need $h$ to be bounded).
$endgroup$
– Paramanand Singh
Jan 27 at 19:25
add a comment |
4
$begingroup$
Try $h(x) = 1/x^2$ for $x ne 0$,$0$ for $x=0$ for a counter example.
$endgroup$
– D.B.
Jan 27 at 4:32
$begingroup$
Why the $2xh+x^2h$ in last equality ? Where does that comes from all of a sudden ?
$endgroup$
– zwim
Jan 27 at 5:07
$begingroup$
@zwim L'hopitale's rule, there was a typo, edited
$endgroup$
– dxdydz
Jan 27 at 18:59
1
$begingroup$
You are not allowed to use L'Hospital's Rule because you don't know anything about $h$ apart from it being bounded. The fraction under limit simplifies to $5+xh(x)$ and this tends to $5$ as $xh(x) to 0$ (and this is where we need $h$ to be bounded).
$endgroup$
– Paramanand Singh
Jan 27 at 19:25
4
4
$begingroup$
Try $h(x) = 1/x^2$ for $x ne 0$,$0$ for $x=0$ for a counter example.
$endgroup$
– D.B.
Jan 27 at 4:32
$begingroup$
Try $h(x) = 1/x^2$ for $x ne 0$,$0$ for $x=0$ for a counter example.
$endgroup$
– D.B.
Jan 27 at 4:32
$begingroup$
Why the $2xh+x^2h$ in last equality ? Where does that comes from all of a sudden ?
$endgroup$
– zwim
Jan 27 at 5:07
$begingroup$
Why the $2xh+x^2h$ in last equality ? Where does that comes from all of a sudden ?
$endgroup$
– zwim
Jan 27 at 5:07
$begingroup$
@zwim L'hopitale's rule, there was a typo, edited
$endgroup$
– dxdydz
Jan 27 at 18:59
$begingroup$
@zwim L'hopitale's rule, there was a typo, edited
$endgroup$
– dxdydz
Jan 27 at 18:59
1
1
$begingroup$
You are not allowed to use L'Hospital's Rule because you don't know anything about $h$ apart from it being bounded. The fraction under limit simplifies to $5+xh(x)$ and this tends to $5$ as $xh(x) to 0$ (and this is where we need $h$ to be bounded).
$endgroup$
– Paramanand Singh
Jan 27 at 19:25
$begingroup$
You are not allowed to use L'Hospital's Rule because you don't know anything about $h$ apart from it being bounded. The fraction under limit simplifies to $5+xh(x)$ and this tends to $5$ as $xh(x) to 0$ (and this is where we need $h$ to be bounded).
$endgroup$
– Paramanand Singh
Jan 27 at 19:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Responding your specific question, you used the fact that $$xcdot h(x)to 0$$
as $x to 0$.
The property that let you conclude this is that zero times bounded goes to zero. More precisely: if $f(x)to 0$ as $xto a$ and $g(x)$ is bounded in a neighborhood of $a$, then
$$f(x)cdot g(x)to 0,$$
as $xto a$. (A detail: none of $f$ or $g$ need to be defined at $x=a$.)
If $g$ is not bounded, the conclusion may or may not be true, as happens when $f(x)=x$ and $g(x)=frac 1 {x^2}$, for which $lim_{xto 0} f(x)=0$, but
$$lim_{xto 0} f(x)cdot g(x)=lim_{xto 0} frac1x=infty.$$
This is possible, since $g(x)$ is not bounded in a neighborhood of $x=0$.
answered Jan 27 at 4:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
add a comment |
$begingroup$
If $h(x)$ is differentiable, then $f(x)=2+5x+x^2h(x)$is differentiable and $$f'(x) = 5+2xh(x)+x^2h'(x)$$
If $h(x)$ and $h'(x)$ are bounded at $x=0$ , then $f'(0) = 5$.
answered Jan 27 at 4:53
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
$begingroup$
Now I noticed that there is no assumption about differentiability of $h(x)$, how do we find $f'(x)$ then?
$endgroup$
– dxdydz
Jan 27 at 5:20
1
$begingroup$
Really no need to add the differentiable condition for $h$, given it is bounded. I think the purpose of this exercise is to unleash the very original definition of the derivative, instead of carrying out algebraic calculation laws.
$endgroup$
– Zhanxiong
Jan 27 at 5:59
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Responding your specific question, you used the fact that $$xcdot h(x)to 0$$
as $x to 0$.
The property that let you conclude this is that zero times bounded goes to zero. More precisely: if $f(x)to 0$ as $xto a$ and $g(x)$ is bounded in a neighborhood of $a$, then
$$f(x)cdot g(x)to 0,$$
as $xto a$. (A detail: none of $f$ or $g$ need to be defined at $x=a$.)
If $g$ is not bounded, the conclusion may or may not be true, as happens when $f(x)=x$ and $g(x)=frac 1 {x^2}$, for which $lim_{xto 0} f(x)=0$, but
$$lim_{xto 0} f(x)cdot g(x)=lim_{xto 0} frac1x=infty.$$
This is possible, since $g(x)$ is not bounded in a neighborhood of $x=0$.
answered Jan 27 at 4:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
add a comment |
$begingroup$
Responding your specific question, you used the fact that $$xcdot h(x)to 0$$
as $x to 0$.
The property that let you conclude this is that zero times bounded goes to zero. More precisely: if $f(x)to 0$ as $xto a$ and $g(x)$ is bounded in a neighborhood of $a$, then
$$f(x)cdot g(x)to 0,$$
as $xto a$. (A detail: none of $f$ or $g$ need to be defined at $x=a$.)
If $g$ is not bounded, the conclusion may or may not be true, as happens when $f(x)=x$ and $g(x)=frac 1 {x^2}$, for which $lim_{xto 0} f(x)=0$, but
$$lim_{xto 0} f(x)cdot g(x)=lim_{xto 0} frac1x=infty.$$
This is possible, since $g(x)$ is not bounded in a neighborhood of $x=0$.
answered Jan 27 at 4:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
add a comment |
$begingroup$
Responding your specific question, you used the fact that $$xcdot h(x)to 0$$
as $x to 0$.
The property that let you conclude this is that zero times bounded goes to zero. More precisely: if $f(x)to 0$ as $xto a$ and $g(x)$ is bounded in a neighborhood of $a$, then
$$f(x)cdot g(x)to 0,$$
as $xto a$. (A detail: none of $f$ or $g$ need to be defined at $x=a$.)
If $g$ is not bounded, the conclusion may or may not be true, as happens when $f(x)=x$ and $g(x)=frac 1 {x^2}$, for which $lim_{xto 0} f(x)=0$, but
$$lim_{xto 0} f(x)cdot g(x)=lim_{xto 0} frac1x=infty.$$
This is possible, since $g(x)$ is not bounded in a neighborhood of $x=0$.
answered Jan 27 at 4:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
Responding your specific question, you used the fact that $$xcdot h(x)to 0$$
as $x to 0$.
The property that let you conclude this is that zero times bounded goes to zero. More precisely: if $f(x)to 0$ as $xto a$ and $g(x)$ is bounded in a neighborhood of $a$, then
$$f(x)cdot g(x)to 0,$$
as $xto a$. (A detail: none of $f$ or $g$ need to be defined at $x=a$.)
If $g$ is not bounded, the conclusion may or may not be true, as happens when $f(x)=x$ and $g(x)=frac 1 {x^2}$, for which $lim_{xto 0} f(x)=0$, but
$$lim_{xto 0} f(x)cdot g(x)=lim_{xto 0} frac1x=infty.$$
This is possible, since $g(x)$ is not bounded in a neighborhood of $x=0$.
answered Jan 27 at 4:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
edited Jan 27 at 5:06
answered Jan 27 at 4:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Jan 27 at 4:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Jan 27 at 4:58
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
4,765118
add a comment |
add a comment |
$begingroup$
If $h(x)$ is differentiable, then $f(x)=2+5x+x^2h(x)$is differentiable and $$f'(x) = 5+2xh(x)+x^2h'(x)$$
If $h(x)$ and $h'(x)$ are bounded at $x=0$ , then $f'(0) = 5$.
answered Jan 27 at 4:53
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
$begingroup$
Now I noticed that there is no assumption about differentiability of $h(x)$, how do we find $f'(x)$ then?
$endgroup$
– dxdydz
Jan 27 at 5:20
1
$begingroup$
Really no need to add the differentiable condition for $h$, given it is bounded. I think the purpose of this exercise is to unleash the very original definition of the derivative, instead of carrying out algebraic calculation laws.
$endgroup$
– Zhanxiong
Jan 27 at 5:59
add a comment |
$begingroup$
If $h(x)$ is differentiable, then $f(x)=2+5x+x^2h(x)$is differentiable and $$f'(x) = 5+2xh(x)+x^2h'(x)$$
If $h(x)$ and $h'(x)$ are bounded at $x=0$ , then $f'(0) = 5$.
answered Jan 27 at 4:53
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
$begingroup$
Now I noticed that there is no assumption about differentiability of $h(x)$, how do we find $f'(x)$ then?
$endgroup$
– dxdydz
Jan 27 at 5:20
1
$begingroup$
Really no need to add the differentiable condition for $h$, given it is bounded. I think the purpose of this exercise is to unleash the very original definition of the derivative, instead of carrying out algebraic calculation laws.
$endgroup$
– Zhanxiong
Jan 27 at 5:59
add a comment |
$begingroup$
If $h(x)$ is differentiable, then $f(x)=2+5x+x^2h(x)$is differentiable and $$f'(x) = 5+2xh(x)+x^2h'(x)$$
If $h(x)$ and $h'(x)$ are bounded at $x=0$ , then $f'(0) = 5$.
answered Jan 27 at 4:53
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
If $h(x)$ is differentiable, then $f(x)=2+5x+x^2h(x)$is differentiable and $$f'(x) = 5+2xh(x)+x^2h'(x)$$
If $h(x)$ and $h'(x)$ are bounded at $x=0$ , then $f'(0) = 5$.
answered Jan 27 at 4:53
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 27 at 4:53
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 27 at 4:53
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Jan 27 at 4:53
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
$begingroup$
Now I noticed that there is no assumption about differentiability of $h(x)$, how do we find $f'(x)$ then?
$endgroup$
– dxdydz
Jan 27 at 5:20
1
$begingroup$
Really no need to add the differentiable condition for $h$, given it is bounded. I think the purpose of this exercise is to unleash the very original definition of the derivative, instead of carrying out algebraic calculation laws.
$endgroup$
– Zhanxiong
Jan 27 at 5:59
add a comment |
$begingroup$
Now I noticed that there is no assumption about differentiability of $h(x)$, how do we find $f'(x)$ then?
$endgroup$
– dxdydz
Jan 27 at 5:20
1
$begingroup$
Really no need to add the differentiable condition for $h$, given it is bounded. I think the purpose of this exercise is to unleash the very original definition of the derivative, instead of carrying out algebraic calculation laws.
$endgroup$
– Zhanxiong
Jan 27 at 5:59
$begingroup$
Now I noticed that there is no assumption about differentiability of $h(x)$, how do we find $f'(x)$ then?
$endgroup$
– dxdydz
Jan 27 at 5:20
$begingroup$
Now I noticed that there is no assumption about differentiability of $h(x)$, how do we find $f'(x)$ then?
$endgroup$
– dxdydz
Jan 27 at 5:20
1
1
$begingroup$
Really no need to add the differentiable condition for $h$, given it is bounded. I think the purpose of this exercise is to unleash the very original definition of the derivative, instead of carrying out algebraic calculation laws.
$endgroup$
– Zhanxiong
Jan 27 at 5:59
$begingroup$
Really no need to add the differentiable condition for $h$, given it is bounded. I think the purpose of this exercise is to unleash the very original definition of the derivative, instead of carrying out algebraic calculation laws.
$endgroup$
– Zhanxiong
Jan 27 at 5:59
add a comment |
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4
$begingroup$
Try $h(x) = 1/x^2$ for $x ne 0$,$0$ for $x=0$ for a counter example.
$endgroup$
– D.B.
Jan 27 at 4:32
$begingroup$
Why the $2xh+x^2h$ in last equality ? Where does that comes from all of a sudden ?
$endgroup$
– zwim
Jan 27 at 5:07
$begingroup$
@zwim L'hopitale's rule, there was a typo, edited
$endgroup$
– dxdydz
Jan 27 at 18:59
1
$begingroup$
You are not allowed to use L'Hospital's Rule because you don't know anything about $h$ apart from it being bounded. The fraction under limit simplifies to $5+xh(x)$ and this tends to $5$ as $xh(x) to 0$ (and this is where we need $h$ to be bounded).
$endgroup$
– Paramanand Singh
Jan 27 at 19:25