Please, I need step by step solution of this two limits. I'd be so greatful for help. please don't mind my stupidity of not knowing L'Hospital rule...

So, this's killing me:

1. lim (x/x-1 - 1/ln x)

x->1

2. lim (x+3/x)^x

x->infinity

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- Jan 30th 2009, 11:07 AMRinoSTkilling limits, please
Please, I need step by step solution of this two limits. I'd be so greatful for help. please don't mind my stupidity of not knowing L'Hospital rule...

So, this's killing me:

1. lim (x/x-1 - 1/ln x)

x->1

2. lim (x+3/x)^x

x->infinity - Jan 30th 2009, 12:18 PMmetlx
2) is there a typo?

is the limit ((x+3)/x)^x?

- Jan 30th 2009, 12:51 PMchiph588@
- Jan 30th 2009, 01:06 PMchiph588@
Note that Since the numerator and denominator approach as approaches , l'Hôpital's rule can be applied.

Notice for the same reasons as above we can apply l'Hôpital's rule again.

Simplifying yields - Feb 3rd 2009, 10:45 AMSoroban
Hello, RinoST!

Let me baby-step through #2 . . .

Quote:

We're expected to know this: .

And we will*hammer*the problem into that form . . .

We have: .

Raise it to the "power 1": .

. . So we have: .

Let

and we have: .

- Feb 3rd 2009, 03:08 PMelliotyang
You can treat [(x+3)/x]^x=y

Of you ln both side, you will get:

ln (y)= x ln[(x+3)/x]

= [(x+3)/x] / (1/x)

Then you can apply L'hopital rule, that is differentiate the numerator and denominator separately

Eventually you will get:

lim{x to infinity} ln y = -3/(x+3)

= 0

lim{x to infinity} y = 1

lim{x to infinity} [(x+3)/x]^x=1

sorry, i do not know how to type the math notation.