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 (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.
lim (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
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 (T-1) + small stuff
log answer = -1 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.
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^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
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.
================================
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.
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
lim (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
thanks
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
N-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.
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