Answer: Limits of Cos(x)/x^2 as x Approaches 0+

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In summary: The Attempt at a SolutionI figured 1/x2 would approach infinity, which and Cos0=1, so the function would approach 1\infty which is \infty, but I was wrong and I got no marks at all for this question. I was trying to find the limit as x-->0+, but I got confused and lost track of what I was doing. Any help at all would be really appreciated. Thanks.
  • #1
Jules18
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Limit as x-->0+

Homework Statement


lim (Cosx)1/(x2)
x-->0+


The Attempt at a Solution



I figured 1/x2 would approach infinity, which and Cos0=1, so the function would approach 1[tex]\infty[/tex] which is [tex]\infty[/tex], but I was wrong and I got no marks at all for this question.

Any help at all would be really appreciated. thanks
 
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  • #2


Jules18 said:

Homework Statement


lim (Cosx)1/(x2)
x-->0+


The Attempt at a Solution



I figured 1/x2 would approach infinity, which and Cos0=1, so the function would approach 1[tex]\infty[/tex] which is [tex]\infty[/tex], but I was wrong and I got no marks at all for this question.

Any help at all would be really appreciated. thanks
[tex][1^{\infty}][/tex] is an indeterminate form, which means that a limit of this type can come out as just about anything.

I would let y = (cos x)1/x2, and then take ln of both sides. Then take the limit of that as x --> 0+. In your notes or text there might be a similar example that you can use as a template.
 
  • #3


yungman said:
This is how I see it:

[tex]\frac{1}{x^2}\rightarrow 0,x\rightarrow0^+[/tex]

[tex]1^0=1[/tex]

Really? I always thought the limit of 1/(x2) would be infinite when x approaches 0.
 
  • #4


yungman said:
This is how I see it:

[tex]\frac{1}{x^2}\rightarrow 0,x\rightarrow0^+[/tex]
This is not true at all. If 1/x2 approaches 0, then x approaches either positive or negative infinity, not zero.
yungman said:
[tex]1^0=1[/tex]
 
  • #5


I'm just guessing, but I think the answer comes out to exp(-1/2).

Take the log of the function to get the variable out of the exponent and find the limit of the log. The answer will be e to the limit you find.
 
  • #6


Jules18 said:
Really? I always thought the limit of 1/(x2) would be infinite when x approaches 0.

Sorry! My mind is not working today, still in the Christmas spirit! Ignor me totally and I'll stay out of this forum today! It's not me, it's my finger...Sorry!
 
  • #7


Jules18 said:

Homework Statement


lim (Cosx)1/(x2)
x-->0+


The Attempt at a Solution



I figured 1/x2 would approach infinity, which and Cos0=1, so the function would approach 1[tex]\infty[/tex] which is [tex]\infty[/tex], but I was wrong and I got no marks at all for this question.

Any help at all would be really appreciated. thanks
[itex]1^\infty[/itex] is NOT 1, it is "indeterminate". If [itex]y= (cos(x))^{1/x^2}[/itex], then [itex]ln(y)= (ln cos(x))/x^2[/itex] which is indeterminate of the form "0/0". Use L'Hopital's rule on that.
 
  • #8


[tex] \ln (y) = \displaystyle \lim_{x\to 0^+} \frac{ln (\cos x)}{x^2} [/tex]

[tex] \ln (y) = \lim_{x\to 0^+} \frac{-\sin x}{2x\cos x} = \lim_{x\to 0^+} \frac{-x+{x^3\over 3!}-\cdots}{2x(1-{x^2\over 2!}\cdots)} =-\frac{1}{2} [/tex]

[tex] y=\frac{1}{\sqrt{e}} [/tex]

Sorry about that.
 
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  • #9


Gregg said:
[tex] \ln (y) = \displaystyle \lim_{x\to 0^+} \frac{1 - \frac{x^2}{2!} +\cdots}{x^2} [/tex]

[tex] \ln (y) = \lim_{x\to 0^+} \frac{-x+\frac{x^3}{3!} +\cdots}{2x} = -\frac{1}{2} [/tex]

[tex] y=\frac{1}{\sqrt{e}} [/tex]

I think the only reason that the "from the right" is there is so the logarithm never has negative values?

Nah. The limit from the left and the right are the same. But your solution has a problem. The numerator is supposed to be log(cos(x))/x^2, not cos(x)/x^2. The latter is infinite.
 
  • #10


Gregg said:
[tex] \ln (y) = \displaystyle \lim_{x\to 0^+} \frac{1 - \frac{x^2}{2!} +\cdots}{x^2} [/tex]
You forgot the log in the numerator.
 
  • #11


Oh yeah, I think it's OK now.
 
  • #12


Gregg said:
[tex] \ln (y) = \displaystyle \lim_{x\to 0^+} \frac{ln (\cos x)}{x^2} [/tex]

[tex] \ln (y) = \lim_{x\to 0^+} \frac{-\sin x}{2x\cos x} = \lim_{x\to 0^+} \frac{-x+{x^3\over 3!}-\cdots}{2x(1-{x^2\over 2!}\cdots)} =-\frac{1}{2} [/tex]

[tex] y=\frac{1}{\sqrt{e}} [/tex]

Sorry about that.

I understand up to = [tex] \lim_{x\to 0^+} \frac{-x+{x^3\over 3!}-\cdots}{2x(1-{x^2\over 2!}\cdots)} [/tex].

I really lost you on that one. Why would that be equal to lim -Sinx/(2xCosx) ??
 
  • #13


Gregg replaced -sin x and cos x by their Maclaurin series. This might be an approach that is more advanced than you are prepared for at this time.
 
  • #14


hm okay well I was given this question on an exam, so I figure my prof had a different approach in mind. Is there a simpler way?
 
  • #15


After taking the log, just use the hospital rule twice.
 

FAQ: Answer: Limits of Cos(x)/x^2 as x Approaches 0+

What is the limit of cos(x)/x^2 as x approaches 0?

The limit of cos(x)/x^2 as x approaches 0 is 1/2. This means that as x gets closer and closer to 0, the value of the expression will approach 1/2.

Why is the limit of cos(x)/x^2 as x approaches 0 important?

The limit of cos(x)/x^2 as x approaches 0 is important because it helps us understand the behavior of the function as x gets closer to 0. It also has many applications in calculus and other areas of mathematics.

How do you find the limit of cos(x)/x^2 as x approaches 0?

To find the limit of cos(x)/x^2 as x approaches 0, you can use the L'Hopital's rule or the Maclaurin series expansion. Alternatively, you can graph the function and observe the behavior as x approaches 0.

What happens if we try to plug in 0 for x in cos(x)/x^2?

If we try to plug in 0 for x in cos(x)/x^2, we will get an indeterminate form of 0/0. This means that the function is undefined at x=0 and we need to use other methods to find its limit at that point.

What are the practical applications of the limit of cos(x)/x^2 as x approaches 0?

The limit of cos(x)/x^2 as x approaches 0 has many practical applications in physics, engineering, and other areas. For example, it can be used to find the maximum height of a projectile, the speed of a race car around a curve, and the stability of a bridge structure.

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