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

[Jules18]

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^\infty which is \infty, but I was wrong and I got no marks at all for this question.

Any help at all would be really appreciated. thanks

 

[Mark44]

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^\infty which is \infty, but I was wrong and I got no marks at all for this question.

Any help at all would be really appreciated. thanks

[1^{\infty}] 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.

 

[Jules18]

This is how I see it:

\frac{1}{x^2}\rightarrow 0,x\rightarrow0^+

1^0=1

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

 

[Mark44]

This is how I see it:

\frac{1}{x^2}\rightarrow 0,x\rightarrow0^+

This is not true at all. If 1/x² approaches 0, then x approaches either positive or negative infinity, not zero.

1^0=1

 

[vela]

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.

 

[yungman]

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!

 

[HallsofIvy]

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^\infty which is \infty, but I was wrong and I got no marks at all for this question.

Any help at all would be really appreciated. thanks

1^\infty is NOT 1, it is "indeterminate". If y= (cos(x))^{1/x^2}, then ln(y)= (ln cos(x))/x^2 which is indeterminate of the form "0/0". Use L'Hopital's rule on that.

 

[Gregg]

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

\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}

y=\frac{1}{\sqrt{e}}

Sorry about that.

 

[Dick]

\ln (y) = \displaystyle \lim_{x\to 0^+} \frac{1 - \frac{x^2}{2!} +\cdots}{x^2}

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

y=\frac{1}{\sqrt{e}}

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.

 

[vela]

\ln (y) = \displaystyle \lim_{x\to 0^+} \frac{1 - \frac{x^2}{2!} +\cdots}{x^2}

You forgot the log in the numerator.

 

[Gregg]

Oh yeah, I think it's OK now.

 

[Jules18]

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

\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}

y=\frac{1}{\sqrt{e}}

Sorry about that.

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

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

 

[Mark44]

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.

 

[Jules18]

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?

 

[vela]

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