Limits Indeterminate forms

Question

$\lim_{x \rightarrow 0^{+}} x^{\frac{1}{\ln x}}$

Answer 1

Let  $y = \lim_{x \to 0^{+} } x^{ 1/\ln x }$

Take $\log$ of both side at base as $e$

$\ln  y = \lim_{x \to 0^{+} } \frac{1}{\ln x} {\ln x}$

$\ln  y = 1$

$\implies y  = e^1$

Answer 2

lim_x->0+ x^(1/lnx)   = lim_x->o+ (e^(1/lnx)*lnx  (since a^b  = e^{b log a)}

 = lim_x->o+ (e^{(lnx/lnx)  =} lim_x->o+ (e^{(1/x)/(1/x)    [applying L'hospital rule since power was in ∞/∞ form]}

^{= e1 = e}

Comments

Look,when ever u will see some power,try to use of log function,cause u have very limited formula for power,but after u take log,power problem will be solved then we can have many ways to solve... And u did correct,but one thing mention the x tends to 0+ in both sides

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I get it !

@Sayantan  put some light on

log y=1   => e

part....how come e ? I am looking for its formula !

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Look, lny means log y base e, so log y =1, y= e^1,which is e.basic log formula