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SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x\to0}\frac{10^{x}1}{x}?\] Examine the onesided 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 onesided limits aren't equivalent, the twosided limit does not exist.

robinfr93
 one year ago
Best ResponseYou've already chosen the best response.0Well my text gives a different answer of 2.59. I stumbled upon the same answer but it says its wrong!! :(

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.0Have you tried putting ln(10) into a calculator? The textbook's answer is probably rounded.

robinfr93
 one year ago
Best ResponseYou've already chosen the best response.0Yeah its 2.3 something. Even if you round it off, there is no way it'll be 2.59

SithsAndGiggles
 one year ago
Best ResponseYou've already chosen the best response.0It might also be the case that the textbook used an approximate method. \[\begin{matrix}\underline{x}& & &\underline{f(x)}\\ 1& & &9\\ 0.1& & &2.58925& & &\text{(probably stopped here)}\\ 0.01& & &2.3293\\ \vdots\end{matrix}\] \[\begin{matrix}\underline{x}& & &\underline{f(x)}\\ 1& & &9\\ 0.1& & &2.58925& & &\text{(again, probably stopped here)}\\ 0.01& & &2.3293\\ \vdots\end{matrix}\]

robinfr93
 one year ago
Best ResponseYou've already chosen the best response.0@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
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