Question

Question #273c3

Answer 1

`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`

Answer 2

`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`