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robinfr93

  • 3 years ago

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

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  1. SithsAndGiggles
    • 3 years ago
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    \[\lim_{x\to0}\frac{10^{|x|}-1}{x}?\] 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
    • 3 years ago
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    Well my text gives a different answer of 2.59. I stumbled upon the same answer but it says its wrong!! :(

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

  4. robinfr93
    • 3 years ago
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    Yeah its 2.3 something. Even if you round it off, there is no way it'll be 2.59

  5. SithsAndGiggles
    • 3 years ago
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    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
    • 3 years ago
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    @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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