Limit of x^cos(1/x) as x approaches 0+ | Calculus Homework Solution

I guessIn summary, the conversation is discussing finding the limit of x^cos(1/x) as x approaches 0 from the positive side. The solution method involves taking the natural logarithm of the expression and using the sandwich theorem and l'hopital's rule. However, it is mentioned that the limit may tend to infinity if cos(1/x) is negative.
  • #1
Snen
3
0

Homework Statement

Homework Equations

The Attempt at a Solution


let y = lim x->0+ x^cos(1/x)
lny = cos(1/x)*lnx = (x*cos(1/x)) * (lnx/x)

x*cos(1/x) = 0 (sandwich theorem)
lnx/x = 0 (l'hopital)
so lny = 0
and y = 1
Is this correct?
 
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  • #2
Snen said:

Homework Statement

Homework Equations

The Attempt at a Solution


let y = lim x->0+ x^cos(1/x)
lny = cos(1/x)*lnx = (x*cos(1/x)) * (lnx/x)

x*cos(1/x) = 0 (sandwich theorem)
lnx/x = 0 (l'hopital)
so lny = 0
and y = 1
Is this correct?
No, that doesn't look right at all.

what happens when ##\cos(1/x)## is negative?
 
  • #3
PeroK said:
No, that doesn't look right at all.

what happens when ##\cos(1/x)## is negative?
PeroK said:
No, that doesn't look right at all.

what happens when ##\cos(1/x)## is negative?
Well then y would tend to infinity but the limit is x-> 0+
 
  • #4
Snen said:
Well then y would tend to infinity but the limit is x-> 0+

He is not asking about what happens to x; he is asking about what happens to x^cos(1/x).
 
  • #5
Ray Vickson said:
He is not asking about what happens to x; he is asking about what happens to x^cos(1/x).
tends to infinity I guess
 

1. What is the limit of "Lim x->0+ (x^cos(1/x))" as x approaches 0 from the right side?

The limit of "Lim x->0+ (x^cos(1/x))" as x approaches 0 from the right side is equal to 1.

2. How do you evaluate "Lim x->0+ (x^cos(1/x))"?

To evaluate "Lim x->0+ (x^cos(1/x))", we can use the squeeze theorem or L'Hopital's rule to simplify the expression and then substitute x=0 to find the limit.

3. Is "Lim x->0+ (x^cos(1/x))" an indeterminate form?

Yes, "Lim x->0+ (x^cos(1/x))" is an indeterminate form of 0 raised to the power of 1, which results in an undefined expression.

4. Can the limit of "Lim x->0+ (x^cos(1/x))" be evaluated from the left side of 0?

No, the limit of "Lim x->0+ (x^cos(1/x))" can only be evaluated from the right side of 0 because the cosine function is not defined for negative values.

5. What is the significance of "Lim x->0+ (x^cos(1/x))" in mathematics?

"Lim x->0+ (x^cos(1/x))" is an example of a limit that cannot be evaluated directly and requires the use of mathematical techniques such as the squeeze theorem or L'Hopital's rule. It also demonstrates the importance of understanding the behavior of functions near points of discontinuity.

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