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

2. 2) is there a typo?
is the limit ((x+3)/x)^x?
$
\lim_{x\to\infty} = (\frac {x+3}{x})^x = \lim_{x\to\infty} = (\frac {x}{x} + \frac {3}{x})^x = \lim_{x\to\infty} = (1 + \frac {3}{x})^x
$

$
(1 + \frac {1}{t})^t = e
$

$
\frac {1}{t} = \frac {3}{x} \Rightarrow x = 3t
$

$
\lim_{x\to\infty} = (1 + \frac {3}{x})^x = \lim_{t\to\infty} = (1 + \frac {1}{t})^{3t} = \lim_{t\to\infty} ((1 + \frac {1}{t})^t)^3 = e^3
$

3. Originally Posted by metlx
2) is there a typo?
if its limit of ((x+3)/x)^x then i know how to solve it. if its (x+3/x)^x then i'm clueless.
If the limit is indeed $\left(x+\frac{3}{x}\right)^x$, it approaches $\infty$ as $x$ approaches $\infty$.

4. Note that $\frac{x}{x-1}-\frac{1}{\ln(x)} = \frac{x\ln(x)-(x-1)}{(x-1)\ln(x)}.$ Since the numerator and denominator approach $0$ as $x$ approaches $1$, l'Hôpital's rule can be applied.

$\lim_{x \to 1} \frac{x\ln(x)-(x-1)}{(x-1)\ln(x)} = \lim_{x \to 1} \frac{\ln(x)}{\ln(x)+\frac{x-1}{x}}$

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

$\lim_{x \to 1} \frac{\ln(x)}{\ln(x)+\frac{x-1}{x}} = \lim_{x \to 1} \frac{\frac{1}{x}}{\frac{1}{x}+\frac{1}{x^2}}$

Simplifying yields $\lim_{x \to 1} \frac{x}{x+1} = \frac{1}{2}$

5. Hello, RinoST!

Let me baby-step through #2 . . .

$\lim_{x\to\infty}\left(\frac{x+3}{x}\right)^x$

We're expected to know this: . $\lim_{z\to\infty}\left(1 + \frac{1}{z}\right)^z \;=\;e$

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

We have: . $\left(\frac{x+3}{x}\right)^x\;=\;\left(\frac{x}{x} + \frac{3}{x}\right)^x \;=\;\left(1 + \frac{3}{x}\right)^x
$

Raise it to the "power 1": . $\bigg[\left(1 + \frac{3}{x}\right)^x\bigg]^{\frac{3}{3}} \;=\;\bigg[\left(1 + \frac{3}{x}\right)^{\frac{x}{3}}\bigg]^3 \;=\;\bigg[\left(1 + \frac{1}{\frac{x}{3}}\right)^{\frac{x}{3}}\bigg]^3$

. . So we have: . $\lim_{x\to\infty}\bigg[\left(1 + \frac{1}{\frac{x}{3}}\right)^{\frac{x}{3}}\bigg]^3$

Let $z \,=\,\frac{x}{3}$

and we have: . $\lim_{z\to\infty}\bigg[\left(1 + \frac{1}{z}\right)^z\bigg]^3 \;=\;\left[\lim_{x\to\infty}\left(1 + \frac{1}{z}\right)^z\right]^3 \;=\;\boxed{e^3}$

6. 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.