Limit : ln-expo

dhiab · #1 $Old$ June 21st, 2009, 10:06 AM

Calculate :

Moo · #2 $Old$ June 21st, 2009, 10:33 AM

Quote:

Originally Posted by dhiab $View Post$

Calculate :

$\lim_{x\to 0} \frac{e^x-e^0}{\ln(x+1)-\ln(0+1)}=\lim_{x\to 0}\frac{e^x-e^0}{x}\cdot\frac{x}{\ln(1+x)-\ln(1+0)}=\frac{f'(0)}{g'(0)}$

where f(x)=e^x and g(x)=ln(1+x)

Krizalid · #3 $Old$ June 21st, 2009, 01:51 PM

the limit is 1 because $\underset{x\to 0}{\mathop{\lim }}\,\frac{e^{x}-1}{x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\ln (1+x)}{x}=1.$

nikhil · #4 $Old$ June 22nd, 2009, 03:48 AM

same can be done using L'Hospital rule (if on putting limit you get a indeterminate form like 0/0 or infinity/infinity, then keep on differentiating both numerator and denominator until you get a fininte limit)

A=lim (e^x-1)/[ln(x+1)]
x->0
on putting x=0 we get
A=(1-1)/ln(1) =0/0
so differentiating both numerator and denominator we get
A=lim (e^x)(x+1)
x->0
now putting x=0
we get A=(e^0)1=1

mr fantastic · #5 $Old$ June 22nd, 2009, 06:32 AM

Quote:

Originally Posted by dhiab $View Post$

Calculate :

Since this question is posted in the PRE-Calculus subforum, I assume a solution that doesn't use calculus is required ....?

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