Question

Find the limit, if it exists;

lim (sin(2x) - 2sqrt(x)*sin(x) + 4x^2) / x
x->0+

Through the use of my TI-89 I found the answer, but I am baffled as to how to get there.

Answer 1

break it up

Answer 2

sin2x / x with go to 2

Answer 3

2sqrt(x)sinx / x goes to zero

Answer 4

as sin x/ x goes to 1 but sqrtx goes to zero..u multiply the limits

Answer 5

4x^2 / x is 4x which goes to zero

Answer 6

so ans is 2 + 0 + 0 = 2

Answer 7

Mmmm....surely sin2x/x=1/2(sin2x/2x), limit 1/2(1) as x goes to 0+

Answer 8

Well, since you get 0/0 (by plugging in) use l'hospital's rule. So:
\[\lim_{x \rightarrow 0^+}\frac{\frac{d}{dx}(\sin(2x)-2\sqrt{x}*\sin(x)+4x^2)}{\frac{d}{dx}(x)}\]
Then you get:
\[\lim_{x \rightarrow 0^+}\frac{2\cos(2x)-2(\frac{1}{2})(\frac{1}{\sqrt{x}})\sin(x)-2\sqrt{x}\cos(x)+8x}{1}\]
Which gives:
\[2\lim_{x \rightarrow 0^+}\cos(2x)-\lim_{x \rightarrow 0^+}\frac{\sin(x)}{\sqrt{x}}-2\lim_{x \rightarrow 0^+}\sqrt{x}\cos(x)+8\lim_{x \rightarrow 0^+}x\]
For the sin term do l'hospital's rule again (0/0)
Giving:
\[-\lim_{x \rightarrow 0^+}\frac{\cos(x)}{\frac{1}{\sqrt{x}}*\frac{1}{2}}\]
Evaluating you get:
2-0-0+0=2.

Answer 9

i apparently wrote all the answers in greek that malevolence found the need to sweat it out...lol

Answer 10

No, I just always attack it that way. I could do the derivatives in my head easily but I wrote it out for demonstrative purposes xP

Answer 11

y do u need derivatives for this? better to explain the fundamental methods

Answer 12

I mean:
\[\lim_{x \rightarrow 0}\frac{\sin(x)}{x}\]
Is done using l'hospital's rule. The derivatives give you the lim x=>0 of cos(x) which is one.

Answer 13

dont use shortcuts where u can avoid em

Answer 14

sin x / x is a standard result

Answer 15

Using l'hospital's xP Either way, equally valid approaches with the same answer. I'm done with calculus and am in further proof-based classes so these exercises aren't as important to me anymore :P

Answer 16

chutiya

Answer 17

the above is without lhospital

Answer 18

thank you myininaya..thats wht i was saying

Answer 19

I mean, either way, that's just as much work as taking some elementary derivatives.

Answer 20

I mean some people see the integral of cos(2x) and have to do a u-sub to realize you divide by 2 to integrate it. I mean, its different methods, same answer.

Answer 21

Thanks for the replies. Malevolence, while the answer and detail is really appreciated, that is way over my head. I have never even heard of l'hospital's rule, so hopefully the professor is not requiring the use of them. I also figured out I had my calculator in degree mode, which was really complicating trying to check these answers.

Anyway, I think Ive got a better grasp on this now. Thanks again and I will definitely return when more questions arise!

Answer 22

Sorry, that is just a standard approach once you have learned the method. Should I have known, I could have explained it just as easily as they did above. I figured instead of just saying: "well sin(x)/x goes to 1" I would give some justification.

Answer 23

malevolence i liked your answer :)

Answer 24

Thank you myininaya :DDDD <3 lol

Answer 25

archetype chances are you will l'hospitals this semester or next semester in cal 2
and you will probably most likely prefer lhospital
it is much easier to use then to think creatively not that lhospital is a mindless process
god i sound mean
im not trying to

Answer 26

than*

Answer 27

Don't make me undo your medal!! ;) Nah, I am willing to soak up any advice that can be given. I am definitely not a natural at math, so I assume most people on here willing to answer my questions will have something to teach me.

I do have three questions regarding your answer Myininaya;

1) Where did the first 2 come from in 2*sin(2x)/2x (third line)

2) How are you able to do 2*sqrt(x) * sin(x)/x (line 3) as opposed to the whole thing being /x (as in line 2)

3) How do you work out sin(2x) / 2x approaches 1 and sin(x)/x approaches 1?

Answer 28

1)i mulitplied by 2/2
2) remember a*c/b=a/b*c or c/b*a
3) remember that sinu/u->1 as u->0

Answer 29

Thank you very much! I did not remember either of those things, but that definitely clears it up.

Answer 30

lol cool