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!
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!