Limit problem

[edumat]

\(\displaystyle lim\,\frac{x\,cos{x}\,+\,e^{-x}}{x²} \,as\,x>0\)

By simply using L'Hôpital's rule twice, I get to \(\displaystyle 1/2\), but the correct result is \(\displaystyle \infty\)

 

[Deleted member 4993]

\(\displaystyle lim\,\frac{x\,cos{x}\,+\,e^{-x}}{x²} \,as\,x>0\)

By simply using L'Hôpital's rule twice, I get to \(\displaystyle 1/2\), but the correct result is \(\displaystyle \infty\)

After taking the limit first time - the result becomes \(\displaystyle \dfrac{1}{0}\)

Since this is not in the "indeterminate" form - you cannot apply L'Hospital's rule.

 

[edumat]

How do I get to \(\displaystyle \infty\) then?

Thanks

 

[Deleted member 4993]

How do I get to \(\displaystyle \infty\) then?

Thanks

\(\displaystyle \dfrac{1}{0} \ = \ ??\)

 

[fcabanski]

You can't use L'Hopital the first time, because the original limit is (0+1)/0. L'Hopital only works when it's 0/0 or inf/inf (plus or minus).

There are many ways to find the limit. Graph the equation. You'll see it approaches infinity from both sides of 0. That's indeterminate, but you can call it infinity. Graphing is generally the best way to begin looking for a limit.

You could also make a table of values approaching 0 from both sides. You'll see that the result becomes larger and larger as you approach 0 from both sides. Again, it will show you infinity.

That's opposed to something like limit as x ---> 0 1/x which approaches infinity from the right, but negative infinity from the left.

 

[HallsofIvy]

Are you saying that you do not know what "limit= \(\displaystyle \infty\)" means?
What does your formula give if x is very close to 0? Why?

 

[lookagain]

You can't use L'Hopital the first time, because the original limit is (0+1)/0. \(\displaystyle \ \ \ \ \) Correct.

L'Hopital only works when it's the fraction in the form of 0/0 or inf/inf (plus or minus).


There are many ways to find the limit. Graph the equation. You'll see it approaches infinity from both sides of 0.

That's indeterminate, \(\displaystyle \ \ \ \ \)No, it is not indeterminate, but it is undefined. **

but you can call it infinity. \(\displaystyle \ \ \ \ \)No, it diverges to (positive) infinity.

Graphing is generally the best way to begin looking for a limit.


You could also make a table of values approaching 0 from both sides.

You'll see that the result becomes larger and larger as you approach 0 from both sides.

Again, it will show you infinity. \(\displaystyle \ \ \ \ \)Doing that will suggest that the limit is \(\displaystyle \ +\infty.\)


That's opposed to something like limit as x ---> 0 \(\displaystyle \ \)of\(\displaystyle \ \ \)1/x, \(\displaystyle \ \ \ \ \)That expression is also undefined at x = 0.

which approaches infinity from the right, but negative infinity from the left.\(\displaystyle \ \ \ \ \)The limit does not exist.

** \(\displaystyle \ \ \ \)"Not every undefined algebraic expression is an indeterminate form.

For example, the expression 1/0 is undefined as a real number but is not indeterminate.

This is because any limit that gives rise to this form will diverge to infinity."

Source: \(\displaystyle \ \ \) http://en.wikipedia.org/wiki/Indeterminate_form

 

[edumat]

Thanks guys!

I just didn't remember that part of limits, but now I see it.

 

[fcabanski]

"That's indeterminate, [FONT=MathJax_Main] [/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Main] [/FONT]No, it is not indeterminate, but it is undefined. ** but you can call it infinity. [FONT=MathJax_Main] [/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Main] [/FONT]No, it diverges to (positive) infinity."

Yes, my mistake.



"You could also make a table of values approaching 0 from both sides. You'll see that the result becomes larger and larger as you approach 0 from both sides. Again, it will show you infinity. [FONT=MathJax_Main] [/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Main] [/FONT][FONT=MathJax_Main] [/FONT]Doing that will suggest that the limit is [FONT=MathJax_Main] [/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]∞[/FONT][FONT=MathJax_Main].[/FONT]"

Please explain that. How would you verify that infinity is the case?

 

[lookagain]

"That's indeterminate, No, it is not indeterminate, but it is undefined. ** but you can call it infinity.
No, it diverges to (positive) infinity."

Yes, my mistake.



"You could also make a table of values approaching 0 from both sides.
You'll see that the result becomes larger and larger as you approach 0
from both sides. Again, it will show you infinity. Doing that will suggest that the limit is [FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]∞[/FONT][FONT=MathJax_Main].[/FONT]"

Please explain that. How would you verify that infinity is the case?

My explanation for the last sentence is that the ever increasing numbers for the outputs in the
table could lend you to believe that the limit is, in fact, \(\displaystyle \ +\infty.\)

However, certain functions can do "weird things" as \(\displaystyle \ x \to \infty \ \) regarding their output values.

Subhotosh Khan is one of the users who told you the form is of the type 1/0.

From that alone, you know the limit is \(\displaystyle \ +\infty.\)

 

[fcabanski]

Thanks for the extra information.