Help pls!! 2 limit questions, feel free to answer both or either one!

[takelight]

a) lim x--> 0⁺ [sin (x)]^x ....................... edited

b) lim x-->infinity (pi/2 - arctan(x))^x


Work attempted: For both these problems I saw they gave indeterminate forms and so I tried using L'hopital's rule by finding derivatives for both. In either case, however, the derivative (found by logarithmic differentiation) gave: (something)(original thing). Thus, I'm back to square one since I still have the original expression in the derivative.

 

[Deleted member 4993]

a) lim x--> 0⁺ (sin (x))^xx

b) lim x-->infinity (pi/2 - arctan(x))^x


Work attempted: For both these problems I saw they gave indeterminate forms and so I tried using L'hopital's rule by finding derivatives for both. In either case, however, the derivative (found by logarithmic differentiation) gave: (something)(original thing). Thus, I'm back to square one since I still have the original expression in the derivative.

L'Hopital's method is NOT applicable here. Plot parts of your functions (using Wolframalfa) and you'll see that the limit is not indeterminate.

 

[takelight]

L'Hopital's method is NOT applicable here. Plot parts of your functions (using Wolframalfa) and you'll see that the limit is not indeterminate.

without plotting how can I know l'hopitals is not applicable???

 

[tkhunny]

without plotting how can I know l'hopitals is not applicable???

Plotting is just a way of investigating the behavior. Feel free to visualize it without plotting.

Please demonstrate how these expressions result in Indeterminate Forms. Introduce a logarithm?

 

[Steven G]

without plotting how can I know l'hopitals is not applicable???

You really should know the answer to your question. Under which conditions is L'Hopital rule applicable? Does your limit satisfy any of the requirements to use L'Hopitals rule?

 

[takelight]

I thought 0^inf or inf^0 or 0^0 were all indeterminate forms. When you plug the limit values into the functions that's exactly what you get.

For a, plug in 0 you get 0^0

for b, 0^infinty

 

[Deleted member 4993]

I thought 0^inf or inf^0 or 0^0 were all indeterminate forms. When you plug the limit values into the functions that's exactly what you get.

For a, plug in 0 you get 0^0

for b, 0^infinty

What do you get when you calculate:

lim x--> 0+ (sin (x)/x) ..................... and

lim x--> 0+ (sin (x)/x)^x ................. and

lim x--> 0+ [(sin (x)/x)^x] * x^(x+1)]

 

[Steven G]

I thought 0^inf or inf^0 or 0^0 were all indeterminate forms. When you plug the limit values into the functions that's exactly what you get.

For a, plug in 0 you get 0^0

for b, 0^infinty

Part a: Your answer is incomplete. Why do you think that you only have to consider part of the expression? Please try again.

 

[takelight]

OK Here's an update:

1) The problem is actually supposed to be lim x--> 0+ (sinx)^x. There is no extra *x there. My bad. I found that I could transform it into e^(x(ln(sinx))), which I thought I could plug in 0, but I still have to deal with the ln(sin(0)) which would be infinity. So I would get 0 times infinity...… ugh.

2) I tried the same thing here, but once again I have to deal with ln(pi/2 - arctan(inf)) which would give ln(0) again so infinity once more. sigh.

I don't know what to do here.

 

[Deleted member 4993]

OK Here's an update:

1) The problem is actually supposed to be lim x--> 0+ (sinx)^x. There is no extra *x there. My bad. I found that I could transform it into e^(x(ln(sinx))), which I thought I could plug in 0, but I still have to deal with the ln(sin(0)) which would be infinity. So I would get 0 times infinity...… ugh.

2) I tried the same thing here, but once again I have to deal with ln(pi/2 - arctan(inf)) which would give ln(0) again so infinity once more. sigh.

I don't know what to do here.

Did you read response #7?

 

[Deleted member 4993]

I thought 0^inf or inf^0 or 0^0 were all indeterminate forms. When you plug the limit values into the functions that's exactly what you get.

For a, plug in 0 you get 0^0

for b, 0^infinty

However, L'Hopitals rule needs a quotient such that the problem turns into 0/0 form. Read:

http://tutorial.math.lamar.edu/Classes/CalcI/LHospitalsRule.aspx

 

[takelight]

Did you read response #7?

I had, but I thought it was written with the wrong knowledge of the first problem *since I wrote the problem incorrectly*. Im guessing only scenarios 1 and 2 are relevant now?
OK so answering that:
1. 0/0 (l'hopitals, right?)
2. um... (0/0)^0, so If I do e^xln(sinx/x). for the ln(sinx/x) since it is in the 0/0 form I can use l'hopitals, and have cosx, then cos(0) = 1, and so I have lim x--> 0+ e^x(ln(1)) and so e^0 which would be 1????? Im wondering If I can use l'hopitals only for some parts of an expression like I did here. like I had to find the lim of sinx/x and so I used l'hopitals only for that bit.

 

[Deleted member 4993]

I had, but I thought it was written with the wrong knowledge of the first problem *since I wrote the problem incorrectly*. Im guessing only scenarios 1 and 2 are relevant now?
OK so answering that:
1. 0/0 (l'hopitals, right?)
2. um... (0/0)^0, so If I do e^xln(sinx/x). for the ln(sinx/x) since it is in the 0/0 form I can use l'hopitals, and have cosx, then cos(0) = 1, and so I have lim x--> 0+ e^x(ln(1)) and so e^0 which would be 1????? Im wondering If I can use l'hopitals only for some parts of an expression like I did here. like I had to find the lim of sinx/x and so I used l'hopitals only for that bit.

1. 0/0 (l'hopitals, right?) ........................ but the limit is 1 after application of L'Hopital's rule.

 

[takelight]

1. 0/0 (l'hopitals, right?) ........................ but the limit is 1 after application of L'Hopital's rule.

Yeah. It is. because you have cosx. I sort of solved it on 2

 

[Steven G]

L'Hopital's method is NOT applicable here. Plot parts of your functions (using Wolframalfa) and you'll see that the limit is not indeterminate.

Prof Khan, can you please update the original post to reflect the correct problem. Otherwise posters will help with the wrong problem. Thanks for you help.

 

[takelight]

ok so please tell me if my reasoning is wrong here:

lim x-->0+ (sinx)^x

e^(x ln(sinx)), so lim x--> 0+ sinx^x is simply e^[whatever x(ln(sinx)) gives me when I check the lim]

so limx-->0+ (x ln(sinx)) gives 0 * infinity, so I try to make it a quotient to use l'hopitals

ln(sinx)/ (x^-1) and now I have infinity/ infinity, thus l'hopitals works.

(cosx/sinx)/(-x^-2) but then this gives me inf/inf again...

 

[Steven G]

ok so please tell me if my reasoning is wrong here:

lim x-->0+ (sinx)^x

e^(x ln(sinx)), so lim x--> 0+ sinx^x is simply e^[whatever x(ln(sinx)) gives me when I check the lim]

so limx-->0+ (x ln(sinx)) gives 0 * infinity, so I try to make it a quotient to use l'hopitals

ln(sinx)/ (x^-1) and now I have infinity/ infinity, thus l'hopitals works.

(cosx/sinx)/(-x^-2) but then this gives me inf/inf again...

Actually it gives you inf/-inf. Be careful! Fortunately it does not matter. So did you try to use L'Hopital's rule again? Is it getting you anywhere?

 

[takelight]

Actually it gives you inf/-inf. Be careful! Fortunately it does not matter. So did you try to use L'Hopital's rule again? Is it getting you anywhere?

Yes. I used it a couple more times and nothing. Why does it not work though, is the method flawed?

Here are my tries:

-(cosx)x²/sinx plug in the 0...…… 0/0

once more.... ((sinx)x² - 2xcosx)/cosx …… plugging in the 0 I get 0/1 which even gives 0 but I know the answer is 1. So where am I messing up?

 

[Steven G]

Let \(\displaystyle y= (sin(x))^x\)

Then \(\displaystyle lny= xln(sin(x))\)

So \(\displaystyle \lim_{x\to 0} lny =\lim_{x\to 0} xln(sin(x))\)

Can you evaluate the right hand limit?

 

[Steven G]

Yes. I used it a couple more times and nothing. Why does it not work though, is the method flawed?

Here are my tries:

-(cosx)x²/sinx plug in the 0...…… 0/0

once more.... ((sinx)x² - 2xcosx)/cosx …… plugging in the 0 I get 0/1 which even gives 0 but I know the answer is 1. So where am I messing up?

L'Hopital's rule is not flawed at all. The rule says that each limit you get after applying L'Hopital's rule will be equal. L'Hopital's rule NEVER said anything about getting a limit that will be easy to compute. Do you understand the difference?