Mathematics

#5: Post edited by $user avatar$ Snoopy‭ · 2024-01-17T12:52:50Z (about 1 year ago)

Copy Link

Raw

Markdown

**Idea.** If $lim_{x \to \infty} f^{'} (x) \neq 0$ , 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 $lim_{x \to \infty} f (x)$ exists. One has a contradiction.
To formalize the idea above, here is the proof.
**Proof.**
Suppose $lim_{x \to \infty} f (x) = L_{0}$ and $lim_{x \to \infty} f^{'} (x) = L$ 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 $x \geq M$ ,
$$
\frac{L}{2}<f'(x)<\frac{3L}{2}.
$$
which implies (for every $x \geq M$ ) 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 $ξ_{x}$ 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 $lim_{x \to \infty} f^{'} (x) \neq 0$ , 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 $lim_{x \to \infty} f (x)$ exists. One has a contradiction.
To formalize the idea above, here is the proof.
**Proof.**
Suppose $lim_{x \to \infty} f (x) = L_{0}$ and $lim_{x \to \infty} f^{'} (x) = L$ 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 $x \geq M$ ,
$$
\frac{L}{2}<f'(x)<\frac{3L}{2}.
$$
which implies (for every $x \geq M$ ) 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 $ξ_{x}$ 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 $user avatar$ Snoopy‭ · 2024-01-14T12:41:45Z (about 1 year ago)

Copy Link

Raw

Markdown

Suppose $lim_{x \to \infty} f (x) = L_{0}$ and $lim_{x \to \infty} f^{'} (x) = L$ 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 $x \geq M$ ,
$$
\frac{L}{2}<f'(x)<\frac{3L}{2}.
$$
which implies (for every $x \geq M$ ) 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 $ξ_{x}$ 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 $lim_{x \to \infty} f^{'} (x) \neq 0$ , 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 $lim_{x \to \infty} f (x)$ exists. One has a contradiction.
To formalize the idea above, here is the proof.
**Proof.**
Suppose $lim_{x \to \infty} f (x) = L_{0}$ and $lim_{x \to \infty} f^{'} (x) = L$ 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 $x \geq M$ ,
$$
\frac{L}{2}<f'(x)<\frac{3L}{2}.
$$
which implies (for every $x \geq M$ ) 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 $ξ_{x}$ 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 $user avatar$ Snoopy‭ · 2024-01-13T23:20:58Z (about 1 year ago)

Copy Link

Raw

Markdown

Suppose $lim_{x \to \infty} f (x) = L_{0}$ and $lim_{x \to \infty} f^{'} (x) = L$ 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 $x \geq M$ ,
$$
\frac{L}{2}<f'(x)<\frac{3L}{2}.
$$
~~which implies (for every $x \geq M$ ),~~
$$
f(x+1)-f(x)=f'(\xi_x)>\frac{L}{2}>0,\tag{1}
$$
where $ξ_{x}$ 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 $lim_{x \to \infty} f (x) = L_{0}$ and $lim_{x \to \infty} f^{'} (x) = L$ 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 $x \geq M$ ,
$$
\frac{L}{2}<f'(x)<\frac{3L}{2}.
$$
which implies (for every $x \geq M$ ) 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 $ξ_{x}$ 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 $user avatar$ Snoopy‭ · 2024-01-13T23:19:33Z (about 1 year ago)

Copy Link

Raw

Markdown

Suppose $lim_{x \to \infty} f (x) = L_{0}$ and $lim_{x \to \infty} f^{'} (x) = L$ 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 $x \geq M$ ,
$$
\frac{L}{2}<f'(x)<\frac{3L}{2}.
$$
~~which implies~~
$$
~~f(x+1)-f(x)=f'(\xi_x)>\frac{L}{2}>0,~~
$$
where $ξ_{x}$ 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 $lim_{x \to \infty} f (x) = L_{0}$ and $lim_{x \to \infty} f^{'} (x) = L$ 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 $x \geq M$ ,
$$
\frac{L}{2}<f'(x)<\frac{3L}{2}.
$$
which implies (for every $x \geq M$ ),
$$
f(x+1)-f(x)=f'(\xi_x)>\frac{L}{2}>0,\tag{1}
$$
where $ξ_{x}$ 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 $user avatar$ Snoopy‭ · 2024-01-13T23:15:17Z (about 1 year ago)

Copy Link

Raw

Markdown

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.

Communities

Post History