What is the Limit of the Square Root Function Using L'Hopital's Rule?

[monet A]

Sorry I had to start a new thread here because the latex was just not generating on the other thread, I don't know if it is my browser or something else but I am having a very difficult day with the tex.

anyhow

$$\lim_{x \rightarrow 0} \frac {\sqrt(5x+9)-3}{x}$$

This gives 0/0 off the bat so apply L'H once I get:

$$\lim_{x \rightarrow 0} \frac {(5x+9).5}{2}$$

Which plugs in nicely to give $\frac {45}{2}$


Did I do that all right?


Hi again can any latex specialists help me. My code definitely says the equation I wrote above is (sqrt(5x+9))-3/x but on my browser there is a red message saying the latex is not valid when I look at the thread and on my preview it is showing me a completely different other tex thing that I wrote two hours ago . I can't seem to avoid this problem, does anyone else see the mix up that I am seeing or is it just my own browser having a hissy fit?

 

[VietDao29]

There's a thread guiding how to use Latex on the General Physics board. Get a look at it.
Are you asking for:
$$\lim_{x \rightarrow 0} \frac{\sqrt{5x + 9} - 3}{x}$$?
If that's what you are asking for, then your answer is wrong. You can recheck your differentiation.
Viet Dao,

 

[monet A]

I am using it right Vietdao, here is my code I have removed the tags so that you can see it.

\lim_{substack{x\rightarrow 0\} \frac {\sqrt{(5x+9)-3}{x}


\lim_{\substack{x\rightarrow 0\} \frac {(5x+9).5}{2}

\frac {45}{2}

 

[VietDao29]

You can click on all Latex images to see its code. Get a look at mine. Obviously, you don't need that '\substack'.
Viet Dao,

 

[monet A]

You can click on all Latex images to see its code. Get a look at mine. Obviously, you don't need that '\substack'.
Viet Dao,

hmm yes I see that but it doesn't explain why my preview kept showing the log function I latexed hours ago fixing the errors was impossible with that going on.

Well I do want to ask you another question but I am not sure if I want another snippy intolerant answer.

Now I know I am an absent minded dumblebum sometimes, but I really love my calculus and I ask this hoping that people will appreciate that some of us did not have the luxury of the complete set of tutorials prior to this point.

Thank you Viet Dao for pointing out that I missed something in my differentiation now that I have found it I need to ask for a little more help.
I keep forgetting what x^(-3/2) translates into, I recall vaguely that it's something like 1/cubrt(x).

Can someone explain to me some way to remember how these (-n/n) indexes translate. Memorising them is getting confusing.

Thankyou

 

[VietDao29]

Okay:
$$a ^ {\frac{m}{n}} = \sqrt[n]{a ^ m}$$
$$a ^ {-m} = \frac{1}{a ^ m}$$
Hope you get it.
Viet Dao,

 

[VietDao29]

Well I do want to ask you another question but I am not sure if I want another snippy intolerant answer.

My posts are snippy? I just want to guide you instead of giving you the answer...
Viet Dao,

 

[monet A]

Thanks

Is that then:

$$\lim_{x \rightarrow 0} \frac{(5x + 9)^{-3/2}*5} {2} = \lim_{x \rightarrow 0} \frac{5} {2\sqrt(5x + 9)^3}$$

?

Sorry if I upset you, really. Darn it its doing the preview thing again, I can't see what I have done wrong with the preview not working. Fixed it.

The working above gives 5/54 is that what you got?

 

[VietDao29]

Hmm, it's incorrect.
It should be:
$$\frac{5}{2\sqrt{5x + 9}}$$
And
$$(x ^ \alpha)' = \alpha x ^{\alpha - 1}$$
Viet Dao,

 

[monet A]

Hmm, it's incorrect.
It should be:
$$\frac{5}{2\sqrt{5x + 9}}$$
And
$$(x ^ \alpha)' = \alpha x ^{\alpha - 1}$$
Viet Dao,

:rofl: Ok maybe I need a headache tablet and a good nights rest.

grr exams making me nutty.

Ok thanks very much Viet Dao for being patient. I don't know why I got -3/2 for a-1 except maybe that I have worn down my brain a bit.

the answer then is 5/6 yes?

 

[VietDao29]

Yes. Anyway, I'll tell you something. To check the lim x -> something answer. Plug the x = some number near something in your function, and if the result is near your answer. Then the chance that you are correct is quite high.
Take this problem an example. Plug x = 0.001 in your calculator. The result will be very near "5 / 6".
Viet Dao,

 

[dextercioby]

Here's an elegant non calculus way to do it

$$\lim_{x\rightarrow 0}\frac{\sqrt{5x+9} -3}{x}\frac{\sqrt{5x+9}+3}{\sqrt{5x+9}+3} =\lim_{x\rightarrow 0}\frac{5x}{x\left(\sqrt{5x+9}+3\right)}=\frac{5}{6}$$

Daniel.

 

[monet A]

Trig limit

Here's an elegant non calculus way to do it

$$\lim_{x\rightarrow 0}\frac{\sqrt{5x+9} -3}{x}\frac{\sqrt{5x+9}+3}{\sqrt{5x+9}+3} =\lim_{x\rightarrow 0}\frac{5x}{x\left(\sqrt{5x+9}+3\right)}=\frac{5}{6}$$

Daniel.

Yeah, thankyou Daniel I deliberately didn't use the conjugates in order to practice L'Hopitals because I had some trouble with it earlier in the day.

Could someone help me with this Trigonometric limit

$$\lim_{t\rightarrow 0}\frac {1-cost^2} {t^2}$$

I realize that I want to manipulate it to look like

$$\lim_{t\rightarrow 0} \frac {1-cost} {t} = 0$$

somewhere, but with trig being a weak point for me I'm a bit put off by the squared variable, I have some idea what to do with trig functions all squared and what to do when they are multiplied by constants but I am a bit lost on the t^2. It might be ridiculously simple, but I need a pointer in the right direction. Thanks

 

[whozum]

$$\lim_{x\rightarrow 0}\frac {1-cost^2} {t^2}$$

If that's $cos^2t$ as I think it is, then the numerator is a simple trig identity. From there its cakework.

edit: Also you might mean the limit as t -> 0

 

[monet A]

$$\lim_{x\rightarrow 0}\frac {1-cost^2} {t^2}$$

If that's $cos^2t$ as I think it is, then the numerator is a simple trig identity. From there its cakework.

Oh gosh I am so embarassed I don't know it and can't seem to find it, is it really that basic?

 

[monet A]

$$\lim_{x\rightarrow 0}\frac {1-cost^2} {t^2}$$

If that's $cos^2t$ as I think it is, then the numerator is a simple trig identity. From there its cakework.

No its cos t^2 not cos^2 t.

edit: Also you might mean the limit as t -> 0

Oh yeah Oops I cut and pasted

:tongue2:

 

[whozum]

Ugh, this is the reason we use parentheses, is it cos(t^2) or cos(t) ^2, as in cos(t) * cos(t)?

 

[monet A]

Oh dear another ooops the actual question I am supposed to answer is

$$\lim_{t\rightarrow 0}\frac {1-cost^2} {t^4}$$

note that the denominator is raised to power 4, since I don't really know what I should be using to solve this I don't know if the t^4 denominator makes any difference, but it probably does .

Still bumbling

 

[monet A]

Ugh, this is the reason we use parentheses, is it cos(t^2) or cos(t) ^2, as in cos(t) * cos(t)?

cos (t^2) The way the difference is shown in my course is by saying

cos^2 t for (cos t)^2
or
cost^2 for cos (t^2)

Sorry to confuse you

 

[whozum]

In that case at a glance I would think the squeeze theorem would suffice.

Also the substitution u = t^2 might make things a bit simpler.

Another idea is multiply by the conjugate.

 

[Jelfish]

I was thinking of ways to make it easier, but just using L'Hospital's rule twice, cancelling the t's until the denominator goes to 1 will get you the answer pretty easily.

 

[monet A]

I did have a look at that last night while I was practising L'hopitals but I think I crunched too far or made a mistake in the numerator because I ended up with -1/6 and using the method that Curious wrote above to check my limit I found that I should reach a limit of zero.

In any case the reason I ask is because I'm 99.99% sure I am asked to revise this because I will be tested on methods of manipulating a trig equation to the form of a basic trig limit and solve, so even when I get the answer by some other method I'll still want to know what I do to the cos (t^2) to split it up into those managable pieces.

Thanks for your thoughts though Jelfish

 

[whozum]

What you'll want to do is multiply from the conjugate, and then use a trig identity to simplify things.

 

[monet A]

What you'll want to do is multiply from the conjugate, and then use a trig identity to simplify things.

OK Whozum I just gave that a go and here's how it turned out for me:

$$\lim_{t\rightarrow 0}\frac {(1-cost^2)(1+cost^2)} {t^4(1+cost^2)} = \frac { 1-cos^2 t^2 } { t^4 (1+cos t^2) } = \frac {sin^2 \Theta} {\Theta^2(1+cos\Theta)} = \frac {sin \Theta} {\Theta} * \frac {sin \Theta} {\Theta} * \frac {1} {1+cos\Theta} = 1*1* \lim_{\Theta\rightarrow 0}\frac {1} {1+cos\Theta} = \frac {1}{2}$$


It looks wrong because it gives an answer I wasn't expecting, I was expecting 0

 

[whozum]

[tex] \frac{1}{2} [/itex] is the correct answer.

 

[monet A]

Thanks for sticking with me Whozum



Monet (swearing I will not take the dog and the kids for a walk both at once ever again.. )