Lim_(xrarr0^+)(1/x)^tanx
We have to apply Euler's identity: e^lnx=x
e^(Lim_(xrarr0^+)ln(1/x)^tanx)
Let's evaluate limit first:
Lim_(xrarr0^+)tanx*ln(1/x)=0*oo
This is type of limit 0*oo which means we can put it in the form oo/oo or 0/0
Lim_(xrarr0^+)(ln(1/x))/(1/(tanx))=Lim_(xrarr0^+)(ln(1/x))/(tanx)^-1=oo/oo
Using L'Hopitals rule:
Lim_(xrarr0^+)(ln(x^-1))/(tanx)^-1
Lim_(xrarr0^+)(1/(x^-1)(-1)x^-2)/(-1*(tanx)^-2*(1/cos^2x))
Lim_(xrarr0^+)(x/x^2cancel((-1)))/(cancel((-1))(tanx)^-2(1/cos^2x))
Lim_(xrarr0^+)(1/x)cos^2x*(tanx)^2
Lim_(xrarr0^+)(cancel(cos^2x)*(sin^2x/cancel(cos^2x)))/x=Lim_(xrarr0^+)(sin^2x)/x=0/0
Using L'Hopitals rule again:
Lim_(xrarr0^+)(sin^2x)/x
Lim_(xrarr0^+)(2sinxcosx)/1=(2*0*1)/1=0/1=0
Answer: e^0=1