calculate limit

Question

$\lim_{x\rightarrow 0} \frac{x^{x}-1}{x}$

Comments

0⁰ is indeterminate form and moreover given expression is not in 0/0 or ∞/∞ form .So we can't apply LH rule directly.

We need to do some transformation

Answer 1

$\lim_{x\rightarrow 0} \frac{x^{x}-1}{x}$ $(0^0 = 1$ and this is of  $\frac{0}{0}$ form)

We apply l'hopital's rule to get

$\lim_{x\rightarrow 0} \frac{\frac{d}{dx}\left(x^{x}\right)}{1}$

To calculate $\lim_{x\rightarrow 0}\frac{d}{dx}(x^{x})$,  put $y=x^{x} $

Taking log both sides,

$\ln y= x \ln x \\ (1/y)(dy/dx)=(1+\ln x) \\ dy=x^x(1+\ln x) dx$

now applying limit we get,

$\lim_{x\rightarrow 0} \frac{d}{dx}\left(x^{x}\right) = \lim_{x\rightarrow 0}x^{x}(1+\ln x) \\= 1.(1+\ln 0) \\= 1+ \ln 0 =\ln 0 \\ \text{Hence, undefined}$

Comments

0 power 0 is 1, not indeterminate form.

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found this just 0^{0 .}is indeterminate form.

http://www.math.fsu.edu/~bellenot/class/f99/cal1/indeterminate.pdf

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seems true.

http://mathforum.org/library/drmath/view/69917.html

Answer 2

First of all,here we can't apply L'Hospital rule because it is not of 0/0 or ∞/∞ form.Here 0⁰ is indeterminate ,so we can't tell its value.

Explanation of why 0⁰  is indeterminate ?

Here 0 is actually tends to zero ,not exactly zero. So it can be either 0⁺  or 0^{- .} So in expression like a^b ( where 'a' and 'b' both can be -ve or +ve ),we can not uniquely determine the value of the expression.That is the reason it is indeterminate.

Now comes to our problem.....

lim_x->0  (x^x -1)/x  =  lim_x->0  (e^xlnx -1)/x -----------------------(A)

Now if x->0  lim_x->0  (xlnx) =  lim_x->0  (lnx/(1/x))   [∞/∞ form]

so applying L'Hospital rule ,we have ,

lim_x->0  (lnx/(1/x)) = lim_x->0  (1/x/(-1/x²)) = 0

so from (A) it is clear that lim_x->0  (e^xlnx -1)/x = (e⁰ -1)/0 => [0/0 form]

Hence now we can apply L'Hospital rule in (A) since is in the form 0/0.

lim_x->0  (e^xlnx -1)/x = lim_x->0  (e^xlnx(1+ lnx))/1   (after differentiation).

Now we can see that the expression lim_x->0  (e^xlnx(1+ lnx))/1 is no longer in indeterminate form. So we can directly put the value of

x->0 and get the result.

Putting x-> 0 in  (e^xlnx(1+ lnx))/1 => e⁰(1 + ln0) = 1(1-∞) = -∞

Hence limiting value of expression is infinite(-ve).