Find out the limit of the following function

In summary, the author has tried different methods to solve a problem but has not been able to find a solution.
  • #1
vkash
318
1
f(0,∞)->R
f(x)=2 [x^(sin(2x)] cos(2x)
find lim(x->0)f(x)=?

I have done all my hits all failed!
can you please tell me how to solve it?
 
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  • #2
Well, tell us what you have tried so we will know what hints will help.
 
  • #3
HallsofIvy said:
Well, tell us what you have tried so we will know what hints will help.

i have reached to this answer! is it correct..

let y = [x^(sin(2x)] cos(2x)
ln(y)= sin(2x)*ln(x)+ln(cos(2x))
lim(x->0) sin(2x)*ln(x)+ln(cos(2x))
lim(x->0) sin(2x)*ln(x) //(as ln(cos(0))=ln(1)=0
lim(x->0)2sin(x)ln(x)
lim(x->0)2 ln(x)/cosec(x)
L Hopital
lim(x->0)-2/(xcosec(x)cot(x))
lim(x->0)-2sin2(x)/xcos(x)
lim(x->0)-2sin2(x)/x
once again L hopital
lim(x->0)-4sin(x)cos(x)+2sin3(x)
=0....(area all steps correct)
ln(y)=0
y=1
;lim(x->0)f(x)=2y=2
is it correct. I doesn't have marking scheme or any thing like this so i can't tell what's correct answer..
 
Last edited:
  • #4
vkash said:
i have reached to this answer! is it correct..

let y = [x^(sin(2x)] cos(2x)
ln(y)= sin(2x)*ln(x)+ln(cos(2x))
lim(x->0) sin(2x)*ln(x)+ln(cos(2x))
lim(x->0) sin(2x)*ln(x) //(as ln(cos(0))=ln(1)=0
lim(x->0)2sin(x)ln(x)
lim(x->0)2 ln(x)/cosec(x)
L Hopital
lim(x->0)-2/(xcosec(x)cot(x))
lim(x->0)-2sin2(x)/xcos(x)
lim(x->0)-2sin2(x)/x
once again L hopital
lim(x->0)-4sin(x)cos(x)+2sin3(x)
=0....(area all steps correct)
ln(y)=0
y=1
;lim(x->0)f(x)=2y=2
is it correct. I doesn't have marking scheme or any thing like this so i can't tell what's correct answer..
The correct answer is 2.
 
  • #5
SammyS said:
The correct answer is 2.
you mean my answer is correct??
OR
you have already done this question and saying me that answer is 2??
 
  • #6
vkash said:
you mean my answer is correct??
OR
you have already done this question and saying me that answer is 2??
The answer is correct.

The step that goes from
lim(x->0) sin(2x)*ln(x)​
to
lim(x->0)2sin(x)ln(x)​
is not correct, if you mean that sin(2x)ln(x) = 2sin(x)ln(x) .
 
  • #7
SammyS said:
The correct answer is 2.

I don't think that's right.
 
  • #8
SammyS said:
The answer is correct.

The step that goes from
lim(x->0) sin(2x)*ln(x)​
to
lim(x->0)2sin(x)ln(x)​
is not correct, if you mean that sin(2x)ln(x) = 2sin(x)ln(x) .

are nahi yar!
no i have just skipped a step..
lim(x->0)sin(2x)ln(x) =lim(x->0)2sin(x)cos(x)ln(x)
cos(0)=1
so
lim(x->0)sin(2x)ln(x) =lim(x->0)2sin(x)cos(x)ln(x)=lim(x->0)2sin(x)ln(x)

so what do you think about it..
 
  • #9
vkash said:
are nahi yar!
no i have just skipped a step..
lim(x->0)sin(2x)ln(x) =lim(x->0)2sin(x)cos(x)ln(x)
cos(0)=1
so
lim(x->0)sin(2x)ln(x) =lim(x->0)2sin(x)cos(x)ln(x)=lim(x->0)2sin(x)ln(x)

so what do you think about it..

Ok so far. I'd double check if you still think the limit is 2.
 
  • #10
I'm pretty sure that [itex]\displaystyle \lim_{x\to0^+}2\,x^{\sin(2x)} \cos(2x)=2\,.[/itex]
 
  • #11
SammyS said:
I'm pretty sure that [itex]\displaystyle \lim_{x\to0^+}2\,x^{\sin(2x)} \cos(2x)=2\,.[/itex]

Yeah, you're right. I've been missing the initial '2' somehow. Sorry.
 
  • #12
:smile:thanks to all of you for helping me...:smile:
 

1. What is the definition of a limit in calculus?

A limit in calculus is a fundamental concept that describes the behavior of a function as its input approaches a certain value. It is denoted by the symbol "lim" and is used to describe the value that a function approaches as the input gets closer and closer to a specific value.

2. How do you find the limit of a function graphically?

To find the limit of a function graphically, you can look at the behavior of the function as the input values get closer and closer to the specific value. If the function approaches a single value as the input approaches the specific value, then that value is the limit of the function.

3. What is the difference between a one-sided limit and a two-sided limit?

A one-sided limit only considers the behavior of the function as the input approaches the specific value from one direction, either the left or the right. A two-sided limit, on the other hand, considers the behavior of the function as the input approaches the specific value from both directions.

4. What are the three conditions for a limit to exist?

The three conditions for a limit to exist are: 1. The function must be defined at the specific value. 2. The function must approach a single value as the input gets closer and closer to the specific value. 3. The behavior of the function as the input approaches the specific value must be the same from both directions for two-sided limits.

5. How do you use algebraic techniques to find the limit of a function?

To use algebraic techniques to find the limit of a function, you can try simplifying the function algebraically or using properties of limits, such as the sum, difference, product, and quotient rules. You can also use L'Hopital's rule or substitution to find the limit of a more complicated function.

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