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January 29, 2015

January 29, 2015

Posted by **Anonymous** on Sunday, February 17, 2008 at 7:28pm.

lim (theta-pi/2)sec(theta)

theta->pi/2

Answer: -1

I am not sure what to do here.

lim (tan(theta))^(theta)

theta->0+

Answer:1

ln(tan(theta))/(1/theta)

L'Hospital's Rule:

lim [(1/(tan(theta))(sec^2(theta))]/(-theta->0+ /x^2]

lim (theta)^2(sin(1/(theta))

theta->oo

Answer:oo

lim (1/h)^2(sin(h))

h->0+

L'Hospital's Rule:

cos(h)/(-2/h)

lim (csc^-1 x)/1/x

x->oo

Answer: 1

I am not sure how 1 can be the answer.

integral 0 to 3 dx/sqrt(9-x^2)

Answer: pi/2

I know this is some form of arcsin(x)->f'(x)=1/sqrt(1-x^2), but I don't know what to do with the 9.

- Calculus -
**Damon**, Sunday, February 17, 2008 at 7:44pmlim (theta-pi/2)sec(theta)

theta->pi/2

Answer: -1

I am not sure what to do here.

++++++++++++++++

(T-pi/2)/cos(T)

cos T = sin(pi/2 - T)

so we have

-(pi/2 -T) / sin (pi/2-T)

as T ---> pi/2 we have sin of small angle and sin(angle) ---->angle + small stuff series in angle^3/3! etc

so in the limit we have

-(pi/2-T)/(pi/2 - T) = 1

- Without series -
**Damon**, Sunday, February 17, 2008 at 8:42pmlim (theta-pi/2)sec(theta)

theta->pi/2

Answer: -1

I am not sure what to do here.

++++++++++++++++

numerator = (T-pi/2)

denominator = cos (T)

d/dT numerator = 1

d/dT denominator = - sin T

as T -->pi/2, sin T ---> 1

so

1/-1

= -1

- Without series -
**Anonymous**, Sunday, February 17, 2008 at 8:46pmthanks

- Without series -

- Without series -
- Calculus -
**Damon**, Sunday, February 17, 2008 at 7:45pmSorry

-1 above

- Calculus -
**Damon**, Sunday, February 17, 2008 at 7:56pmlim (tan(theta))^(theta)

theta->0+

Answer:1

ln(tan(theta))/(1/theta)

L'Hospital's Rule:

lim [(1/(tan(theta))(sec^2(theta))]/(-theta->0+ /x^2]

============================

(tan T)^T as T--->0+

log answer = T ln tan T

log answer = T ln ( T + small stuff)

log answer = T (T-1) + small stuff

log answer = -1 T --->0

e^0 = 1

- without series - what a mess -
**Damon**, Sunday, February 17, 2008 at 9:02pmMy way sure is easier !

ln answer = T ln tan T

= ln tan T/1/T

numerator = ln tan T

denominator = 1/T

d/dT numerator = (cos T/sin T)(1/cos^2 T)

= 1/(cos T sin T)

but

sin T cos T = .5 sin 2T

d/dT denominator = -1/T^2

as T --> 0 sin 2T ---> 2 T

so

we have

1/T / -1/T^2 = -T ---> 0

e^0 = 1

- without series - what a mess -
- Calculus -
**Damon**, Sunday, February 17, 2008 at 7:58pmI am seeing a pattern here. Use a Taylor series expansion for your functions.

- Calculus -
**Damon**, Sunday, February 17, 2008 at 8:03pmlim (theta)^2(sin(1/(theta))

theta->oo

Answer:oo

lim (1/h)^2(sin(h))

h->0+

L'Hospital's Rule:

cos(h)/(-2/h)

++++++++++++++++++++++++++++++++++=

T^2 sin(1/T) as T--> oo

again sin A ---> A -A^3/3!..... for small angle

so we have

T^2/ (1/T) = T ^3 for big T

that gets very big as T gets moderately large.

- Calculus -
**Damon**, Sunday, February 17, 2008 at 9:12pmnumerator = sin (1/T)

denominator = 1/T^2

d/dT numerator = -(1/T^2 )cos (1/T)

d/dt denominator = -2T/T^4 = - 2/T^3

ratio = 2 T cos (1/T)

cos 0 = 1

so

ratio = 2 T which is oo as T ---> oo

- Calculus -
- Calculus -
**Damon**, Sunday, February 17, 2008 at 8:14pmlim (csc^-1 x)/1/x

x->oo

Answer: 1

I am not sure how 1 can be the answer.

==========================

numerator = csc^-1 x

denominator = (1/x)

d/dx numerator = -1/x sqrt(x^2-1)

d/dx denominator = -1/x^2

as x gets big we have -1/x^2 for the numerator and -1/x^2 for the denominator

- Calculus -
**Damon**, Sunday, February 17, 2008 at 8:19pmintegral 0 to 3 dx/sqrt(9-x^2)

Answer: pi/2

I know this is some form of arcsin(x)->f'(x)=1/sqrt(1-x^2), but I don't know what to do with the 9.

================================

let x = 3 sin t

dx = 3 cos t dt

so

3 cos t dt /sqrt(9 - 9 sin^2 t)

3 cos t dt / 3 sqrt (1-sin^2 t)

but 1 - sin^2 = cos ^2

You can take it from there.

- Calculus (for damon) -
**Anonymous**, Sunday, February 17, 2008 at 8:23pmabout the taylor expansion. I haven't learned that yet. Is there an alternative way?

- Calculus (for damon) -
**Damon**, Sunday, February 17, 2008 at 8:34pmHave you had any series even in algebra 2 ?

- Calculus (for damon) -
- Calculus -
**Anonymous**, Sunday, February 17, 2008 at 8:39pmyes, but that is a long time ago

- Calculus -
**Count Iblis**, Sunday, February 17, 2008 at 9:08pmN-th order Taylor expansion:

f(x+h) = f(x) + h f'(x) + h^2/2 f''(x) +h^3/3! f'''(x) + ...+ h^N/N! f^(N)(x) +

O(h^N+1)

Assuming that f is continuously differentiable N times. O(h^(N+1)) means a term proportional to h^(N+1) in the limit h ---> 0.

A non-rigorous proof goes as follows. Just assume that f(x+h) can be expressed as a power series in h:

f(x + h) = A + B h + C h^2 + ...

Then take the limit h --->0 of both sides to find A. Take the first derivative and the limit h ---> 0 to find B, the second derivative and the limit h ---> 0 to find C, etc.

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