limit help

**davecs77** · Jul 14th 2007, 11:38 AM

I have the limit as x goes to 1 (1/lnx - x/x-1)
I am seeing a lot of these limits, with two different fractions. Is a good strategy finding a common denonometer and making it into one fraction? Then use Lhosipatals rule if it applies to find the limit? That is what im trying to do but I am not getting the answer in the book.

**galactus** · Jul 14th 2007, 12:19 PM

$\lim_{x\rightarrow{1}}[\frac{1}{ln(x)}-\frac{x}{x-1}]$

Rewrite as:

$\lim_{x\rightarrow{1}}\frac{-xln(x)+x-1}{xln(x)-ln(x)}$

L'Hopital gives:

$\lim_{x\rightarrow{1}}\frac{-xln(x)}{x-1+xln(x)}$

L'Hopital again:

$\lim_{x\rightarrow{1}}\frac{-ln(x)-1}{ln(x)+2}=\frac{-1}{2}$

**davecs77** · Jul 14th 2007, 12:44 PM

Originally Posted by galactus

$\lim_{x\rightarrow{1}}[\frac{1}{ln(x)}-\frac{x}{x-1}]$

Rewrite as:

$\lim_{x\rightarrow{1}}\frac{-xln(x)+x-1}{xln(x)-ln(x)}$

L'Hopital gives:

$\lim_{x\rightarrow{1}}\frac{-xln(x)}{x-1+xln(x)}$

L'Hopital again:

$\lim_{x\rightarrow{1}}\frac{-ln(x)-1}{ln(x)+2}=\frac{-1}{2}$

How did you rewrite that fraction? I dont see that.

**galactus** · Jul 14th 2007, 01:03 PM

That's just what you would do if you were subtracting. Basic algebra.

$\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}$

**davecs77** · Jul 14th 2007, 01:15 PM

Originally Posted by galactus

$\lim_{x\rightarrow{1}}[\frac{1}{ln(x)}-\frac{x}{x-1}]$

Rewrite as:

$\lim_{x\rightarrow{1}}\frac{-xln(x)+x-1}{xln(x)-ln(x)}$

L'Hopital gives:

$\lim_{x\rightarrow{1}}\frac{-xln(x)}{x-1+xln(x)}$

L'Hopital again:

$\lim_{x\rightarrow{1}}\frac{-ln(x)-1}{ln(x)+2}=\frac{-1}{2}$

When you take the derivative of (-xln(x) + x - 1)/xln(x) - ln(x) I am getting
(-x - ln(x))/x+ln(x)-(1/x)
What am i doing wrong? EDIT NVM i forgot a 1
lol.

**galactus** · Jul 14th 2007, 01:27 PM

Yeah, the biggest trouble with calc is the algebra. It can be tricky and easy to 'flub up'.

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