Post History
#5: Post edited
by
Snoopy
·
2024-01-17T12:52:50Z (4 months ago)
- **Idea.** If $\displaystyle \lim_{x\to\infty}f'(x)\ne 0$, then for large enough $x$, $f'(x)$ is bounded away from $0$, which implies by the mean value theorem that $|f(x+1)-f(x)|$ is bounded away from $0$. But $|f(x+1)-f(x)|$ must also be small for large $x$ since $\displaystyle \lim_{x\to\infty}f(x)$ exists. One has a contradiction.
- To formalize the idea above, here is the proof.
- **Proof.**
- Suppose $\displaystyle \lim_{x\to\infty}f(x)=L_0$ and $\displaystyle \lim_{x\to\infty}f'(x)=L$ for some real numbers $L_0$ and $L$. We show that $L=0$.
- For the sake of argument, suppose $L>0$. Then it follows from the limit definition that there exists some positive real number $M>0$ such that for every $x\ge M$,
- $$
- \frac{L}{2}<f'(x)<\frac{3L}{2}.
- $$
- which implies (for every $x\ge M$) by the [mean value theorem](https://en.wikipedia.org/wiki/Mean_value_theorem),
- $$
- f(x+1)-f(x)=f'(\xi_x)>\frac{L}{2}>0,\tag{1}
- $$
where $\xi_x$ is some real number depends on $x$.- On the other hand, since $\lim_{x\to\infty}f(x)=L_0$, for large enough $x$ that is greater than $M$, one has
- $$
- |f(x+1)-f(x)|<|f(x+1)-L_0|+|f(x)-L_0|<\frac{L}{2},
- $$
- which contradicts (1).
- Similarly, one can argue that $L<0$ is impossible.
- **Idea.** If $\displaystyle \lim_{x\to\infty}f'(x)\ne 0$, then for large enough $x$, $f'(x)$ is bounded away from $0$, which implies by the mean value theorem that $|f(x+1)-f(x)|$ is bounded away from $0$. But $|f(x+1)-f(x)|$ must also be small for large $x$ since $\displaystyle \lim_{x\to\infty}f(x)$ exists. One has a contradiction.
- To formalize the idea above, here is the proof.
- **Proof.**
- Suppose $\displaystyle \lim_{x\to\infty}f(x)=L_0$ and $\displaystyle \lim_{x\to\infty}f'(x)=L$ for some real numbers $L_0$ and $L$. We show that $L=0$.
- For the sake of argument, suppose $L>0$. Then it follows from the limit definition that there exists some positive real number $M>0$ such that for every $x\ge M$,
- $$
- \frac{L}{2}<f'(x)<\frac{3L}{2}.
- $$
- which implies (for every $x\ge M$) by the [mean value theorem](https://en.wikipedia.org/wiki/Mean_value_theorem),
- $$
- f(x+1)-f(x)=f'(\xi_x)>\frac{L}{2}>0,\tag{1}
- $$
- where $\xi_x$ is some real number between $x$ and $x+1$.
- On the other hand, since $\lim_{x\to\infty}f(x)=L_0$, for large enough $x$ that is greater than $M$, one has
- $$
- |f(x+1)-f(x)|<|f(x+1)-L_0|+|f(x)-L_0|<\frac{L}{2},
- $$
- which contradicts (1).
- Similarly, one can argue that $L<0$ is impossible.
#4: Post edited
by
Snoopy
·
2024-01-14T12:41:45Z (4 months ago)
- Suppose $\displaystyle \lim_{x\to\infty}f(x)=L_0$ and $\displaystyle \lim_{x\to\infty}f'(x)=L$ for some real numbers $L_0$ and $L$. We show that $L=0$.
- For the sake of argument, suppose $L>0$. Then it follows from the limit definition that there exists some positive real number $M>0$ such that for every $x\ge M$,
- $$
- \frac{L}{2}<f'(x)<\frac{3L}{2}.
- $$
- which implies (for every $x\ge M$) by the [mean value theorem](https://en.wikipedia.org/wiki/Mean_value_theorem),
- $$
- f(x+1)-f(x)=f'(\xi_x)>\frac{L}{2}>0,\tag{1}
- $$
- where $\xi_x$ is some real number depends on $x$.
- On the other hand, since $\lim_{x\to\infty}f(x)=L_0$, for large enough $x$ that is greater than $M$, one has
- $$
- |f(x+1)-f(x)|<|f(x+1)-L_0|+|f(x)-L_0|<\frac{L}{2},
- $$
- which contradicts (1).
- Similarly, one can argue that $L<0$ is impossible.
- **Idea.** If $\displaystyle \lim_{x\to\infty}f'(x)\ne 0$, then for large enough $x$, $f'(x)$ is bounded away from $0$, which implies by the mean value theorem that $|f(x+1)-f(x)|$ is bounded away from $0$. But $|f(x+1)-f(x)|$ must also be small for large $x$ since $\displaystyle \lim_{x\to\infty}f(x)$ exists. One has a contradiction.
- To formalize the idea above, here is the proof.
- **Proof.**
- Suppose $\displaystyle \lim_{x\to\infty}f(x)=L_0$ and $\displaystyle \lim_{x\to\infty}f'(x)=L$ for some real numbers $L_0$ and $L$. We show that $L=0$.
- For the sake of argument, suppose $L>0$. Then it follows from the limit definition that there exists some positive real number $M>0$ such that for every $x\ge M$,
- $$
- \frac{L}{2}<f'(x)<\frac{3L}{2}.
- $$
- which implies (for every $x\ge M$) by the [mean value theorem](https://en.wikipedia.org/wiki/Mean_value_theorem),
- $$
- f(x+1)-f(x)=f'(\xi_x)>\frac{L}{2}>0,\tag{1}
- $$
- where $\xi_x$ is some real number depends on $x$.
- On the other hand, since $\lim_{x\to\infty}f(x)=L_0$, for large enough $x$ that is greater than $M$, one has
- $$
- |f(x+1)-f(x)|<|f(x+1)-L_0|+|f(x)-L_0|<\frac{L}{2},
- $$
- which contradicts (1).
- Similarly, one can argue that $L<0$ is impossible.
#3: Post edited
by
Snoopy
·
2024-01-13T23:20:58Z (4 months ago)
- Suppose $\displaystyle \lim_{x\to\infty}f(x)=L_0$ and $\displaystyle \lim_{x\to\infty}f'(x)=L$ for some real numbers $L_0$ and $L$. We show that $L=0$.
- For the sake of argument, suppose $L>0$. Then it follows from the limit definition that there exists some positive real number $M>0$ such that for every $x\ge M$,
- $$
- \frac{L}{2}<f'(x)<\frac{3L}{2}.
- $$
which implies (for every $x\ge M$),- $$
- f(x+1)-f(x)=f'(\xi_x)>\frac{L}{2}>0,\tag{1}
- $$
- where $\xi_x$ is some real number depends on $x$.
- On the other hand, since $\lim_{x\to\infty}f(x)=L_0$, for large enough $x$ that is greater than $M$, one has
- $$
- |f(x+1)-f(x)|<|f(x+1)-L_0|+|f(x)-L_0|<\frac{L}{2},
- $$
- which contradicts (1).
- Similarly, one can argue that $L<0$ is impossible.
- Suppose $\displaystyle \lim_{x\to\infty}f(x)=L_0$ and $\displaystyle \lim_{x\to\infty}f'(x)=L$ for some real numbers $L_0$ and $L$. We show that $L=0$.
- For the sake of argument, suppose $L>0$. Then it follows from the limit definition that there exists some positive real number $M>0$ such that for every $x\ge M$,
- $$
- \frac{L}{2}<f'(x)<\frac{3L}{2}.
- $$
- which implies (for every $x\ge M$) by the [mean value theorem](https://en.wikipedia.org/wiki/Mean_value_theorem),
- $$
- f(x+1)-f(x)=f'(\xi_x)>\frac{L}{2}>0,\tag{1}
- $$
- where $\xi_x$ is some real number depends on $x$.
- On the other hand, since $\lim_{x\to\infty}f(x)=L_0$, for large enough $x$ that is greater than $M$, one has
- $$
- |f(x+1)-f(x)|<|f(x+1)-L_0|+|f(x)-L_0|<\frac{L}{2},
- $$
- which contradicts (1).
- Similarly, one can argue that $L<0$ is impossible.
#2: Post edited
by
Snoopy
·
2024-01-13T23:19:33Z (4 months ago)
- Suppose $\displaystyle \lim_{x\to\infty}f(x)=L_0$ and $\displaystyle \lim_{x\to\infty}f'(x)=L$ for some real numbers $L_0$ and $L$. We show that $L=0$.
- For the sake of argument, suppose $L>0$. Then it follows from the limit definition that there exists some positive real number $M>0$ such that for every $x\ge M$,
- $$
- \frac{L}{2}<f'(x)<\frac{3L}{2}.
- $$
which implies- $$
f(x+1)-f(x)=f'(\xi_x)>\frac{L}{2}>0,- $$
- where $\xi_x$ is some real number depends on $x$.
On the other hand, since $\lim_{x\to\infty}f(x)=L_0$, for large enough $x$, one has- $$
- |f(x+1)-f(x)|<|f(x+1)-L_0|+|f(x)-L_0|<\frac{L}{2},
- $$
which is a contradiction.- Similarly, one can argue that $L<0$ is impossible.
- Suppose $\displaystyle \lim_{x\to\infty}f(x)=L_0$ and $\displaystyle \lim_{x\to\infty}f'(x)=L$ for some real numbers $L_0$ and $L$. We show that $L=0$.
- For the sake of argument, suppose $L>0$. Then it follows from the limit definition that there exists some positive real number $M>0$ such that for every $x\ge M$,
- $$
- \frac{L}{2}<f'(x)<\frac{3L}{2}.
- $$
- which implies (for every $x\ge M$),
- $$
- f(x+1)-f(x)=f'(\xi_x)>\frac{L}{2}>0,\tag{1}
- $$
- where $\xi_x$ is some real number depends on $x$.
- On the other hand, since $\lim_{x\to\infty}f(x)=L_0$, for large enough $x$ that is greater than $M$, one has
- $$
- |f(x+1)-f(x)|<|f(x+1)-L_0|+|f(x)-L_0|<\frac{L}{2},
- $$
- which contradicts (1).
- Similarly, one can argue that $L<0$ is impossible.
#1: Initial revision
by
Snoopy
·
2024-01-13T23:15:17Z (4 months ago)
Suppose $\displaystyle \lim_{x\to\infty}f(x)=L_0$ and $\displaystyle \lim_{x\to\infty}f'(x)=L$ for some real numbers $L_0$ and $L$. We show that $L=0$.
For the sake of argument, suppose $L>0$. Then it follows from the limit definition that there exists some positive real number $M>0$ such that for every $x\ge M$,
$$
\frac{L}{2}<f'(x)<\frac{3L}{2}.
$$
which implies
$$
f(x+1)-f(x)=f'(\xi_x)>\frac{L}{2}>0,
$$
where $\xi_x$ is some real number depends on $x$.
On the other hand, since $\lim_{x\to\infty}f(x)=L_0$, for large enough $x$, one has
$$
|f(x+1)-f(x)|<|f(x+1)-L_0|+|f(x)-L_0|<\frac{L}{2},
$$
which is a contradiction.
Similarly, one can argue that $L<0$ is impossible.