# Limit

• Apr 17th 2009, 11:05 AM
Apprentice123
Limit
Consider a function $f$ that $|f(x) - Ax| \leq e^{-x}$, for any number A $\in$ R. Calculate:

$lim_x{ \to +\infty} \frac{f(x) - f(1)}{x-1}$
• Apr 17th 2009, 12:41 PM
running-gag
Hi

$\frac{|f(x) - f(1)|}{|x-1|} = \frac{|(f(x) - Ax) + (Ax - A) + (A - f(1))|}{|x-1|}$

$\frac{|f(x) - f(1)|}{|x-1|} \leq \frac{|f(x) - Ax| + |Ax - A| + |A - f(1)|}{|x-1|}$

Spoiler:

$\frac{|f(x) - f(1)|}{|x-1|} \leq \frac{|f(x) - Ax|}{|x-1|} + \frac{|Ax - A|}{|x-1|} + \frac{|A - f(1)|}{|x-1|}$

$\frac{|f(x) - f(1)|}{|x-1|} \leq \frac{e^{-x}}{|x-1|} + |A| + \frac{e^{-1}}{|x-1|}$

And

$\lim_{x \rightarrow +\infty} \left(\frac{e^{-x}}{|x-1|} + |A| + \frac{e^{-1}}{|x-1|}\right) = |A|$

$\forall A \in \mathbb{R} ~ \lim_{x \rightarrow +\infty}\frac{|f(x) - f(1)|}{|x-1|} \leq |A|$

Therefore $\lim_{x \rightarrow +\infty}\frac{|f(x) - f(1)|}{|x-1|} = 0$

It would be better if someone could confirm ! (Wink)

• Apr 17th 2009, 03:05 PM
Apprentice123
Thank you men