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$f(x)$ is a differentiable function on the real line such that $\underset{x \to \infty=}{\lim} f(x) =1$ and $\underset{x \to \infty=}{\lim} f’(x) =\alpha$. Then

  1. $\alpha$ must be $0$
  2. $\alpha$ need not be $0$, but $\mid \alpha \mid <1$
  3. $\alpha >1$
  4. $\alpha < -1$
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$$f(n+1)-f(n)=\frac{f(n+1)-f(n)}{(n+1)-n}=f'(c_n)\;\;,\;\;\;c_n\in (n,\,n+1)$$

using the MVT (Lagrange's Theorem) for differentiable functions.

From this it follows that  $\;c_n\xrightarrow[n\to\infty]{}\infty\;$, and thus

$$\alpha=\lim_{n\to\infty}f'(c_n)=\lim_{n\to\infty}\left(f(n+1)-f(n)\right)=1-1=0$$

Refer from following link:

https://math.stackexchange.com/a/1257166/952652

 

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