Question

Solve for Lim (10^|X| - 1)/x
x->0

Answer 1

\[\lim_{x\to0}\frac{10^{|x|}-1}{x}?\]
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.

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

@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