Diffferent Result with LHopitals Rule!

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expertinsights

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Jan 11, 2010
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Everyone,

Little stumbled upon this:

Lim x -> -inf (10*x-3*cos(3*x))/(2*x)

When I solve this limit WITHOUT L'Hopital rule I get 5. However I applied LHopital's rule and I get a range of values!

Any insight into why I get different results would be helpful. Thanks.
 
5 is correct.

\(\displaystyle \lim_{x\to -\infty}\frac{10x-3cos(3x)}{2x}\)

That is because when we differentiate and get \(\displaystyle \frac{10+9sin(3x)}{2}\) does not exist, nor is it \(\displaystyle \pm {\infty}\)
 
thank u galactus.

I guess my question is why is that I am getting a different result when I apply the LHopitals rule.

After I apply the LHopital rule I get: (10 + 9Sin3x)/2 and when we do the -inf limit we get a diffferent value set!

Why is this so...is it that we can't apply the LHopital rule here? ( I think we should be able to as it evaluates to an indefinite form)

Thanks!
 
I edited my post. Check above. After we differentiate, what we get does not exist.

If the limit were askin for x to 0 instead of infinity, then we are in business. But look at the graph and you can see as \(\displaystyle x\to \pm{\infty}\), then limit does not exist.

Try this one as well:

\(\displaystyle \lim_{x\to {\infty}}\frac{x+sin(2x)}{x}\)
 
The Marqui is of no avail here, ergo the squeeze play to the rescue.

\(\displaystyle -1 \ \le \ cos(3x) \ \le1, \ -3 \ \le \ -3cos(3x) \ \le \ 3\)

\(\displaystyle 10x-3 \le \ 10x-3cos(3x) \ \le \ 10x+3, \ \frac{10x+3}{2x} \ \le \ \frac{10x-3cos(3x)}{2x} \ \le \ \frac{10x-3}{2x}, \ x \ is \ -\)

\(\displaystyle Hence, \ \lim_{x\to-\infty}\frac{10x+3}{2x} \ \le \ \lim_{x\to-\infty}\frac{10x-3cos(3x)}{2x} \ \le \ \lim_{x\to-\infty}\frac{10x-3}{2x}\)

\(\displaystyle Now. \ the \ Marqui \ comes \ into \ play, \ to \ wit: \ 5 \ \le \ \lim_{x\to-\infty}\frac{10x-3cos(3x)}{2x} \ \le \ 5\)

\(\displaystyle Therefore, \ only \ one \ way \ to \ go \ - \ \lim_{x\to-\infty}\frac{10x-3cos(3x)}{2x} \ = \ 5\)

\(\displaystyle Note: \ sin(x) \ and/or \ cos(x) \ as \ x \ approaches \ \pm \ \infty \ diverges \ by \ oscillation, \ hence \ no \ can\)

\(\displaystyle \ use \ the \ Marqui.\)

\(\displaystyle Also \ note: \ As \ x \ goes \ to \ \pm \ \infty, \ -3cos(3x) \ pales \ to \ nothing, \ leaving \ only \ \frac{10x}{2x}.\)
 
Just to clarify, L'Hopital's rule does always work; it is a proved theorem. The hypothesis requires that

\(\displaystyle \lim_{x \to a} \frac{f'(x)}{g'(x)}\) exists.

"Exists" may be taken in context to include \(\displaystyle \pm \infty\). Be weary of oscillating functions and L'Hopital's rule (sometimes it may be applied, sometimes not).
 
Maybe I should have elaborated. I did not mean it that way. I just meant in some cases, one has to be careful. As in the above case.

Most are used to L'Hopital working right out of the box. That is, taking the derivatives and then it works. But in some cases, like this one, the

derivatives fail to exist and we run into a small 'snag'.

As we know, sin and cos oscillate between -1 and 1 and when x--inf, we hit the 'snag' I mentioned. That's all.

Sorry for the misunderstanding and confusion. If L'Hopital does not work so well it is because of something like the above. I did not mean that it does not work...period.

Glenn, you do not have to call me 'someone'. You can call me Cody :D

BTW, here is a fun limit that come from the idea that all indeterminate forms like \(\displaystyle 0^{0}, \;\ {\infty}^{0}, \;\ 1^{\infty}\) have the

value of 1 because anything to the 0 power is 1 and 1 to any power is 1.

\(\displaystyle 0^{0}, \;\ {\infty}^{0}, \;\ 1^{\infty}\) are not powers of numbers, but limits.

\(\displaystyle \lim_{x\to {+ \infty}}x^{\frac{ln(a)}{1+ln(x)}}\)

\(\displaystyle \lim_{x\to 0}(x+1)^{\frac{ln(a)}{x}}\)
 
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