Question #273c3

2 Answers
Dec 23, 2017

lim_(x->0^+)(1/x)^tan(x)
=exp(lim_(x->0^+)tan(x)ln(1/x))
=exp(-lim_(x->0^+)tan(x)ln(x))

Since tan(x)=sin(x)/cos(x),
=exp(-lim_(x->0^+)ln(x)/(cos(x)/sin(x)))

Use L'Hôpital's rule:
=exp(lim_(x->0^+)(1/x)/(1/sin^2(x)))

Rearrange to
=exp(lim_(x->0^+)sin(x)/x*sin(x))

Then,
=exp(lim_(x->0^+)(sin(x)/x)*lim_(x->0^+)(sin(x)))

Using the fact that lim_(x->0)sin(x)/x=1,
=exp(lim_(x->0^+)sin(x))
=exp(0)
=1

Dec 23, 2017

Lim_(xrarr0^+)(1/x)^tanx=e^0=1

Explanation:

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