1>Lim (ln(ln(x))/ln(x)
x->inf
(a)-1 (b) 0 (c) 1 (d) inf

2>An ideal gas satisfies the equation PV = RT, where P is the pressure, V is the volume, R is a constant, T is the temperature, and n is the number of moles of the gas. State the rate of change of the temperature with respect to time in terms of P, V, R, and n.
(a) Vdp/dt +pDv/dt )/ndR/dt (b)0 (c) (vdp/dt +p DV/dt)/nR (d) none

3>An ideal gas satisfies the equation PV = RT, where P is the pressure, V is the volume, R is a constant, and T is the temperature. At a certain time, the temperature is maintained constant, the pressure P = 100 lb/in2 and is increasing at 4 lb/in2•sec. At what rate is the volume changing when it is 60 in3?
(a) -2.4 (b) 2.4 (c)0 (d) none

4>Lim sqrt(x+1)/ln(x+1)
x->0
(a) limit does not exist (b)0 (c)1 (d) 1/2

5>Lim 2x ln(x^2)
x->0
(a)inf (b) -1(c) 0 (d) -1/2

6>Lim (1+5/x) ^x
x->inf
(a)5 (b)1 (c) e (d) e^5

2. Originally Posted by bobby77
1>Lim (ln(ln(x))/ln(x)
x->inf
(a)-1 (b) 0 (c) 1 (d) inf
The answer can't be $\infty$, as this is the same as:

$
\lim_{x\rightarrow \infty} \frac{\ln(x)}{x}
$

which is $0$, as L'Hopitals rule will verify (and numerical
experiment confirm).

RonL

3. [QUOTE=bobby77]
2>An ideal gas satisfies the equation PV = RT, where P is the pressure, V is the volume, R is a constant, T is the temperature, and n is the number of moles of the gas. State the rate of change of the temperature with respect to time in terms of P, V, R, and n.
(a) Vdp/dt +pDv/dt )/ndR/dt (b)0 (c) (vdp/dt +p DV/dt)/nR (d) none

First the ideal gas equation is:

$
PV=nRT
$

where $n$ is the number of moles, and $R$ is the universal gas
constant, and $P,\ V$ and $T$ are the Pressure Volume
and Temprature respectivly.

Now:

$
\frac{d}{dt}PV=P\frac{dV}{dt}+\frac{dP}{dt}V=nR \frac{dT}{dt}
$

So:

$
\frac{dT}{dt}=\left( \frac{P\frac{dV}{dt}+\frac{dP}{dt}V}{nR} \right)
$

RonL

4. Originally Posted by bobby77
3>An ideal gas satisfies the equation PV = RT, where P is the pressure, V is the volume, R is a constant, and T is the temperature. At a certain time, the temperature is maintained constant, the pressure P = 100 lb/in2 and is increasing at 4 lb/in2•sec. At what rate is the volume changing when it is 60 in3?
(a) -2.4 (b) 2.4 (c)0 (d) none
From (2) we get:

$
\frac{d}{dt}PV=P\frac{dV}{dt}+\frac{dP}{dt}V=R \frac{dT}{dt}
$

( $n$ had been absorbed into [math[R[/tex] in this case)

Now we are told that:

$
\frac{dT}{dt}=0,\ \ \frac{dP}{dt}=4,\ \ P=100,\ \mbox{and}\ V=60
$

so:

$
P\frac{dV}{dt}=-V\frac{dP}{dt}
$
,

form which it follows that

$
\frac{dV}{dt}=-\ \frac{4 \times 60}{100}=-2.4
$

RonL

5. Originally Posted by bobby77
4>Lim sqrt(x+1)/ln(x+1)
x->0
(a) limit does not exist (b)0 (c)1 (d) 1/2

5>Lim 2x ln(x^2)
x->0
(a)inf (b) -1(c) 0 (d) -1/2
get your calculator out and just try these for x=10, 100, 1000. This should
be sufficient to identify what these limits are.

RonL

6. Originally Posted by bobby77
6>Lim (1+5/x) ^x
x->inf
(a)5 (b)1 (c) e (d) e^5
The advice given for problems (4) and (5) also applies here. But this
is usefull to know:

$
\lim_{x \rightarrow \infty}(1+1/x)^x=e
$

So putting $y=zx$ gives for fixed $z$:

$
\lim_{y \rightarrow \infty}(1+z/y)^{y/z}=e
$

and so:

$
\lim_{y \rightarrow \infty}(1+z/y)^{y}=e^z
$

So here the answer is (d)

RonL