continuous function problem

[shirkay]

Hi, having trouble solving this problem: sinx to the power of alpha multiplied by cos1/x all divided by x^3

f(x) = [sin^α(x)*cos(1/x)]/x^3 , x≠0

0 , x≠0

Now they ask:which values of alpha make the function continuous at x=0?

I am trying to check what happens when alpha is larger than 0.

lim f(x)= 0/0 -----------> Tried using L'hopital:
x--> 0+

[α*sin^(α-1)x*cosx*cos(1/x) + sin^α(x)*-sin(1/x)*(-1/x^2)]/2x^3 -----> limit is still 0/0... using L'hopital again would be crazy, I'm sure there is a different way or maybe I approached the question incorrectly from the beginning.

I know I need to check the limit of both when x---> 0+ and x----> 0- and find out if the limits are the same but I am stuck finding the limit itself.
Help will be much appreciated

• :grin:

 

[Deleted member 4993]

Hi, having trouble solving this problem: sinx to the power of alpha multiplied by cos1/x all divided by x^3

f(x) = [sin^α(x)*cos(1/x)]/x^3 , x≠0

0 , x≠0

Now they ask:which values of alpha make the function continuous at x=0?

I am trying to check what happens when alpha is larger than 0.

lim f(x)= 0/0 -----------> Tried using L'hopital:
x--> 0+

[α*sin^(α-1)x*cosx*cos(1/x) + sin^α(x)*-sin(1/x)*(-1/x^2)]/2x^3 -----> limit is still 0/0... using L'hopital again would be crazy, I'm sure there is a different way or maybe I approached the question incorrectly from the beginning.

I know I need to check the limit of both when x---> 0+ and x----> 0- and find out if the limits are the same but I am stuck finding the limit itself.
Help will be much appreciated

• :grin:

What is:

\(\displaystyle \lim_{x\to 0}\dfrac{sin^3(x)}{x^3}\)

What is:

\(\displaystyle \lim_{x\to 0}\left [sin(x) * cos(\dfrac{1}{x})\right ]\)

Use wolframalpha.com to guide you.

 

[DrPhil]

Hi, having trouble solving this problem: sinx to the power of alpha multiplied by cos1/x all divided by x^3

f(x) = [sin^α(x)*cos(1/x)]/x^3 , x≠0

0 , x≠0 I assume this is supposed to be f(x) = 0 when x=0

Now they ask:which values of alpha make the function continuous at x=0?

I am trying to check what happens when alpha is larger than 0.

lim f(x)= 0/0 -----------> Tried using L'hopital:
x--> 0+

[α*sin^(α-1)x*cosx*cos(1/x) + sin^α(x)*-sin(1/x)*(-1/x^2)]/2x^3 -----> limit is still 0/0... using L'hopital again would be crazy, I'm sure there is a different way or maybe I approached the question incorrectly from the beginning.

I know I need to check the limit of both when x---> 0+ and x----> 0- and find out if the limits are the same but I am stuck finding the limit itself.
Help will be much appreciated

• :grin:

\(\displaystyle \displaystyle \lim_{x\to 0}\dfrac{\sin^{\alpha}(x) \cos(1/x)}{x^3}\)

First, note that cos(1/x) will oscillate rapidly between +1 and -1, so you can never know what the sign of the numerator is. Then leave it out of your calculation.

Second, If l'Hospital gives you 0/0, you CAN do it again. Eventually either the nth derivative of numerator or denominator will not go to zero.

Now concentrate on the difference between the powers of \(\displaystyle \alpha\) in the numerator and \(\displaystyle 3\) in the denominator. Maybe you could replace \(\displaystyle \alpha\) with
\(\displaystyle \alpha = 3 + (\alpha - 3)\)
What do you know about \(\displaystyle [\sin(x)/x]^3\) ?