L’ hospital rule

[Joana]

Please help me evaluate the following limit:
Lim x->0+ [(x + SinX)] ^(1/x)

here’s what what I did:
lny = 1/x ln(x+SinX)
=> lim x->0+ lny = lim x->0+ 1/x ln(x+SinX)

This is where I’m stuck since the limit doesn’t yield the intermediate form (ln(0)=undefined).

I also tried -1 =< SinX =< 1 (Squeeze theorem)
=> lim x->0+ [(x-1)] ^ (1/x) =< Lim x->0+ [(x + SinX)] ^(1/x) =< Lim x->0+ [(x+1)] ^ (1/x).
I then tried to find the limit of [(x-1)] ^ (1/x) and [(x+1)] ^ (1/x) by setting both sides to logarithm (just like the first one I did) which turned into:

lim x->0+ [ ln(x-1) ] / x (L’ Hospital’s rule doesn’t work for this one since ln (0-1)/0 = ln (-1)/ 0 and ln(-1) doesn’t exist). I’m stuck here (

lim x->0+ [ ln(x+1) ] / x (this works fine).

Thanks in advance!

 

[Steven G]

Yes, ln(0) is undefined but why are you mentioning this?

You want to compute lim x->0+ [ln(x+SinX)], that is you want to know what ln(x+SinX) gets closet to as x gets closer to 0+! This has nothing to do with what ln(0) equals.

Please try again.

 

[Joana]

Yes, ln(0) is undefined but why are you mentioning this?

You want to compute lim x->0+ [ln(x+SinX)], that is you want to know what ln(x+SinX) gets closet to as x gets closer to 0+! This has nothing to do with what ln(0) equals.

Please try again.

but ln (0 + sin0) = ln(0)?

 

[Steven G]

Yes, but x will never be 0! x will get extremely close to 0 but not 0! Just look at the ln(x) graph and tell me what the y-value is approaching as x gets closer and closer to 0? Substitution does not always work!

 

[Dr.Peterson]

Please help me evaluate the following limit:
Lim x->0+ [(x + SinX)] ^(1/x)

here’s what what I did:
lny = 1/x ln(x+SinX)
=> lim x->0+ lny = lim x->0+ 1/x ln(x+SinX)

This is where I’m stuck since the limit doesn’t yield the intermediate form (ln(0)=undefined).

This is confusing because you used the wrong word. You meant to say "an indeterminate form", meaning 0/0 or inf/inf. In fact, you didn't say yet that you were trying to apply L'Hopital!

You're right if you mean that L'Hopital doesn't apply to this form. In fact, ln(x+sin(x))/x has the form -inf/0, which clearly approaches -infinity. You don't need L'Hopital.