$f(x)$ is a differentiable function on the real line such that $ \lim_{x\to \infty } f(x) =1 $ and $ \lim_{x\to \infty } f'(x) = s $ .Then
- $s$ should be $0$
- $s$ need not be $0$ but $|s| < 1$
- $s > 1$
- $s < -1$
Because $f(x)$ is bounded need not mean it can neither be increasing nor decreasing.So the derivative needs not be $0$.
$\endgroup$ 35 Answers
$\begingroup$By Mean Value theorem we have the following equation $$f(x + 1) - f(x) = f'(\xi)$$ where $x < \xi < x + 1$. The LHS of the above equation tends to $1 - 1 = 0$ as $x \to \infty$ and RHS tends to $s$ as $x \to \infty$. Hence $s$ must be $0$.
$\endgroup$ 1 $\begingroup$Since $$\lim_{x\to\infty}f'(x)=s$$
That means for all $\epsilon>0$, there exists $R$ and $\delta>0$ s.t. $$\bigg|\frac{f(x+h)-f(x)}{h}-s\bigg|<\epsilon$$ whenever $x>R$ and $h<\delta$.
In other words, $$s-\epsilon<\frac{f(x+h)-f(x)}{h}<s+\epsilon$$
Now suppose $s\ne 0$, then set $\epsilon<|s|/2$ and $h=\delta/2$, we have $$\bigg|f(x+\delta/2)-f(x)\bigg|>\frac{|s|\delta}{4}$$
However, since $$\lim_{x\to\infty}f(x)=1$$ we know that there exist $R'$ so that $$\bigg|f(x)-1\bigg|<\frac{|s|\delta}{8}$$ whenever $x>R'$.
Now for $x>\max(R,R')$, we have $$\bigg|f(x+\delta/2)-f(x)\bigg| \le |f(x+\delta/2)-1|+|f(x)-1|<\frac{|s|\delta}{4}$$ which is a contradiction.
Hence $s=0$.
$\endgroup$ $\begingroup$If the derivative does not approach zero at infinity, the function value will continue to change (non-zero slope). Since we know the function is a constant, the derivative must go to zero.
Just pick an $|s| < 1,$ and draw what happens as you do down the real line. If $s \neq 0,$ the function can't remain a constant.
$\endgroup$ 5 $\begingroup$Assume that the tangent of $f(x)$ is given by $$g(x)=mx+k$$ where m is the slope. But $m=\frac{dy}{dx}=f'(x)$
So $$g(x)=f'(x)x+k$$ As $x\rightarrow \infty$, we know that the asymptote is the horizontal line $y=1$, and it follows that $$\lim_{x\rightarrow \infty} g(x)=1\\\lim_{x\rightarrow \infty}(f'(x)x+k) =1$$
For this to be true, we must have an indeterminate form concerning the product $f'(x)x$. But $\lim_{x\rightarrow \infty} x=\infty$, so this can only happen if $\lim_{x\rightarrow \infty} f'(x)=0$
Thus, $s=0$.
$\endgroup$ 1 $\begingroup$Using the definition of derivative, $$f'(x)_{\text{at } x\to \infty} = \lim_{x \to \infty} \lim_{h\to 0+} \frac{f(x+h) - f(h)}{h} $$
But, since $x \to \infty$, $(x + h) \to \infty$
So, $$\begin{align} f'(x) &= \frac{f(\to \infty) - f (\to \infty)}{h} \\ &= \frac{1-1}{h} \\ &= \frac 0 h \\ &= 0 \end{align}$$
So, $s=0$.
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