robinfr93 2 years ago Solve for Lim (10^|X| - 1)/x x->0

1. SithsAndGiggles

$\lim_{x\to0}\frac{10^{|x|}-1}{x}?$ Examine the one-sided limits. $\lim_{x\to0^+}\frac{10^{|x|}-1}{x}$ As x approaches 0 from the right, you'll be considering only positive values of x, which means |x|=x. $\lim_{x\to0^+}\frac{10^{x}-1}{x}=\frac{0}{0}\\ \text{Apply L'Hopital's rule}\\ \lim_{x\to0^+}\frac{\ln(10)\cdot10^{x}}{1}=\frac{\ln(10)}{1}=\ln(10)$ $\lim_{x\to0^-}\frac{10^{|x|}-1}{x}$ As x approaches 0 from the left, you'll be considering negative values of x, which means |x|=-x. $\lim_{x\to0^-}\frac{10^{-x}-1}{x}=\frac{0}{0}\\ \text{Apply L'Hopital's rule}\\ \lim_{x\to0^+}\frac{-\ln(10)\cdot10^{-x}}{1}=-\frac{\ln(10)}{1}=-\ln(10)$ Since the one-sided limits aren't equivalent, the two-sided limit does not exist.

2. robinfr93

Well my text gives a different answer of 2.59. I stumbled upon the same answer but it says its wrong!! :(

3. SithsAndGiggles

Have you tried putting ln(10) into a calculator? The textbook's answer is probably rounded.

4. robinfr93

Yeah its 2.3 something. Even if you round it off, there is no way it'll be 2.59

5. SithsAndGiggles

It might also be the case that the textbook used an approximate method. $\begin{matrix}\underline{x}& & &\underline{f(x)}\\ 1& & &9\\ 0.1& & &2.58925& & &\text{(probably stopped here)}\\ 0.01& & &2.3293\\ \vdots\end{matrix}$ $\begin{matrix}\underline{x}& & &\underline{f(x)}\\ -1& & &-9\\ -0.1& & &-2.58925& & &\text{(again, probably stopped here)}\\ -0.01& & &-2.3293\\ \vdots\end{matrix}$

6. robinfr93

@SithsAndGiggles Hmmm.. Could be.. Anyways thanx for helping.. I really appreciate it.. just for that I'll give you a medal, would had knighted you to Sir. SithsAndGiggles if had gotten a sure answer.. Anwys thanx again