Calculus
posted by Anonymous .
Below are the 5 problems which I had trouble in. I can't seem to get the answer in the back of the book. Thanks for the help!
lim (thetapi/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(9x^2)
Answer: pi/2
I know this is some form of arcsin(x)>f'(x)=1/sqrt(1x^2), but I don't know what to do with the 9.

lim (thetapi/2)sec(theta)
theta>pi/2
Answer: 1
I am not sure what to do here.
++++++++++++++++
(Tpi/2)/cos(T)
cos T = sin(pi/2  T)
so we have
(pi/2 T) / sin (pi/2T)
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/2T)/(pi/2  T) = 1 
lim (thetapi/2)sec(theta)
theta>pi/2
Answer: 1
I am not sure what to do here.
++++++++++++++++
numerator = (Tpi/2)
denominator = cos (T)
d/dT numerator = 1
d/dT denominator =  sin T
as T >pi/2, sin T > 1
so
1/1
= 1 
thanks

Sorry
1 above 
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]
============================
(tan T)^T as T>0+
log answer = T ln tan T
log answer = T ln ( T + small stuff)
log answer = T (T1) + small stuff
log answer = 1 T >0
e^0 = 1 
My 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 
I am seeing a pattern here. Use a Taylor series expansion for your functions.

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)
++++++++++++++++++++++++++++++++++=
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. 
numerator = 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 
lim (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^21)
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 
integral 0 to 3 dx/sqrt(9x^2)
Answer: pi/2
I know this is some form of arcsin(x)>f'(x)=1/sqrt(1x^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 (1sin^2 t)
but 1  sin^2 = cos ^2
You can take it from there. 
about the taylor expansion. I haven't learned that yet. Is there an alternative way?

Have you had any series even in algebra 2 ?

yes, but that is a long time ago

Nth 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 nonrigorous 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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