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It's pretty sad, but I've been working on this math problem for a couple of hours, now. Still scouring my Calculus textbook (Calculus Concepts and Contexts by James Stewart,) class notes, and Math SE, but I can't seem to find an example similar to this that involves natural logs where $x$ is approaching $1$.

Basically, I am trying to work with the following expression so that it no longer has an indeterminate form $\frac{0}{0}$ in order to be able to ultimately use limit laws to simplify the expression.

I believe I've been understanding the concepts behind the previous few problems, but when the natural log is involved in this particular expression's denominator, I'm getting stuck as the denominator appears to continually equal $0$. (And my goal is to somehow find a way to have this denominator not equal $0$.)

Here is the expression I am working with:

$$ \lim_{x \to 1} \frac{4ln(x) + 2ln(x^3)}{ln(x)-ln(\sqrt{x})} $$

Here is what I've been trying:

$$ \lim_{x \to 1} \frac{4ln(x) + 2ln(x^3)}{ln(x)-ln(\sqrt{x})} $$

$$ \lim_{x \to 1} \frac{4ln(x) + 3 * 2ln(x)}{ln(x)-ln(\sqrt{x})} $$

$$ \lim_{x \to 1} \frac{10ln(x)}{ln(x)-ln(\sqrt{x})} * \frac{ln(x)+ln(\sqrt{x})}{ln(x)+ln(\sqrt{x})} $$

$$ \lim_{x \to 1} \frac{10ln^2(x) + 10(ln(x)+ln(\sqrt{x}))}{ln^2(x)-ln(x)} $$

From there, no matter what I seem to do (by trying to cancel factors out, etc.) due to $x$ approaching $1$, and the natural log of $1$ being 0... I will continue to get a denominator of $0$.

Am I missing some sort of secret related to natural log rules? Or (if not,) am I missing an important step within my existing process?

Edit:

Here is what I ended up using:

$$ \lim_{x \to 1} \frac{4ln(x) + 2ln(x^3)}{ln(x)-ln(\sqrt{x})} $$

$$ \lim_{x \to 1} \frac{4ln(x) + 3 * 2ln(x)}{ln(x)-ln(\sqrt{x})} $$

$$ \lim_{x \to 1} \frac{4ln(x) + 6ln(x)}{ln(x)-ln(x^\frac{1}{2})} $$

$$ \lim_{x \to 1} \frac{4ln(x) + 6ln(x)}{ln(x)-\frac{1}{2}ln(x)} $$

$$ \lim_{x \to 1} \frac{10ln(x)}{\frac{1}{2}ln(x)} $$

$$ \lim_{x \to 1} \frac{10}{\frac{1}{2}} $$

$$ \frac{\lim_{x \to 1} 10}{\lim_{x \to 1} \frac{1}{2}} $$

$$ \frac{10}{\frac{1}{2}} $$

$$ 20 $$

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2  
Your treating the logarithm like a mononomial, multiplying the numerator and denominator by its conjuguate wont make things simpler, just simplify $\ln(\sqrt{x})=\frac{1}{2}\ln(x)$ –  Ethan Apr 17 '13 at 6:32
4  
You have used $\log x^3=3\log x$ --- good! Now, what can you do, along the same lines, with $\log\sqrt x$? –  Gerry Myerson Apr 17 '13 at 6:33
    
@GerryMyerson I think I've got it, now... for some reason I don't remember learning about $\sqrt{x} = x^\frac{1}{2}$ in my past classes; glad to know about it now, though! –  summea Apr 17 '13 at 7:08

2 Answers 2

up vote 1 down vote accepted

$$ \lim_{x \to 1} \frac{4\ln(x) + 2\ln(x^3)}{\ln(x)-\ln(\sqrt{x})}=\lim_{x \to 1} \frac{4\ln(x) + 6\ln(x)}{\frac{1}{2}\ln(x)}=\frac{4+6}{\frac{1}{2}}=20 $$

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Thanks for your time; I'm definitely a slow learner, but I appreciate the help! –  summea Apr 17 '13 at 7:02

Use the L Hopital's rule and you will get your answer in a single step.

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I've been seeing this around online; we haven't quite gotten to that rule (in class) but I'm looking forward to using it, someday! Thanks~ –  summea Apr 17 '13 at 7:03
2  
@summea Soon I will edit it to make you more clear. –  srijan Apr 17 '13 at 7:06

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