How to Simplify Tricky Calculus Limits?

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In summary, the conversation discusses the process of simplifying certain limits, specifically those involving square roots and trigonometric functions. The conversation also mentions the use of L'Hopital's rule and the importance of being patient and polite when seeking help from others.
  • #1
cracker
37
0
Need help with these dam limits!

Ok how do I simplify this ****? Like I know what to do with the problem but IDK how to simplify it look.

1. Lim [(3+x)^(1/2) - (3)^(1/2)] / x
x>0

2. Lim (sinx) / 5x
x>0

3. Lim (sin^2x) / x
x>0

4. Lim (1-cosh)^2 / h
x>0

Answers are: Got them from the back of the book
1. (3)^(1/2) / 6
2. (1/5)
3. 0
4. 0
 
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  • #2
come on peeps! I should know this but the teach did not explain how to simplify these only the ones like x^2 - 3x + 5, ect. he did not mentions any with square roots :(
 
  • #3
Do you know L'Hopital's rule?
 
  • #4
hhmm... what? Hospital?
 
  • #5
Well, find out--it's in your book and online.
 
  • #6
0rthodontist said:
Well, find out--it's in your book and online.

If your not going be any help then maby you should just not comment on my thread.
 
  • #7
Everyone here is offering their time and help for free and it may take some time for someone to reply to a post, so please have a little patience.

For the first one, the common trick is to multiply the numerator and denominator by the cojugate of the numerator, ie: multiply the numerator and denominator by [itex] \sqrt{3+x} + \sqrt{3}[/itex]

The second and third are easy if you know ahead of time that [tex]\lim_{x \rightarrow 0}\frac{sin(x)}{x} = 1 [/tex].

For the fourth do you mean [tex] \lim_{h \rightarrow 0}{\frac{(1-cos(h))^2}{h}[/tex] ?
 
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  • #8
nocturnal said:
Everyone here is offering their time and help for free and it may take some time for someone to reply to a post, so please have a little patience.

For the first one, the common trick is to multiply the numerator and denominator by the cojugate of the numerator, ie: multiply the numerator and denominator by [itex] \sqrt{3+x} + \sqrt{3}[/itex]

The second and third are easy if you know ahead of time that [tex]\lim_{x \rightarrow 0}\frac{sin(x)}{x} = 1 [/tex].

OMFG dude thx! I forgot about all that :D! And I'm try to be more patient but that guy was annoying.
 
  • #9
Hmm... wait I don't understand the first one you said
 
  • #10
He said to multiply the expression by 1 written as

[tex]\frac{ \sqrt{3+x} + \sqrt{3}}{\sqrt{3+x} + \sqrt{3}}[/tex]

You'll see that those limits can all be done in the same direct way once you learn about l'Hospital's rule. (Yes, Hospital)
 
  • #11
cracker said:
OMFG dude thx! I forgot about all that :D!
You're welcome :smile:
cracker said:
If your not going be any help then maby you should just not comment on my thread.
cracker said:
I'm try to be more patient but that guy was annoying.
You'll find that people here are very willing to help and do so generously for free. The very least you can do is be polite and show some appreciation. It's okay to say that you don't understand a hint or suggestion or that its not helping you, but comments like these are likely to turn people away, not to mention they're just plain immature and disrespectful.
 
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  • #12
quasar987 said:
He said to multiply the expression by 1 written as

[tex]\frac{ \sqrt{3+x} + \sqrt{3}}{\sqrt{3+x} + \sqrt{3}}[/tex]

You'll see that those limits can all be done in the same direct way once you learn about l'Hospital's rule. (Yes, Hospital)

K thx. Dam my teachers an *****. How the **** was I supposed to know that.
 
  • #13
How were you supposed to know what? When it was suggested before that you multiply numerator and denominator by [itex]\sqrt{3+x}+\sqrt{3}[/itex] your response was "I forgot about all that". Then you said "How the **** was I supposed to know that." It is, after all, basic algebra and you are in a Calculus course. It seems to me your "

The person who was so "irritating" to you suggested a perfectly valid method of doing this problem, it just happened that you have not yet learned it. Since you gave no indication that you had tried any method at all, there was no indication of what you did or did not know. He had no way of knowing that you weren't familiar with L'Hopital's method.

Finally, I note that there was exactly 10 minutes between your first post and your "come on peeps!". I would expect a response to a question in one or two days. It's not like you were paying a tutor.

(Oh, and lose the "valley girl" vocabulary- it tends to put people off.)How were you supposed to know what? When it was suggested before that you multiply numerator and denominator by [itex]\sqrt{3+x}+\sqrt{3}[/itex] your response was "I forgot about all that". Then you said "How the **** was I supposed to know that." It is, after all, basic algebra and you are in a Calculus course. It seems to me your "

The person who was so "irritating" to you suggested a perfectly valid method of doing this problem, it just happened that you have not yet learned it. Since you gave no indication that you had tried any method at all, there was no indication of what you did or did not know. He had no way of knowing that you weren't familiar with L'Hopital's method.

Finally, I note that there was exactly 10 minutes between your first post and your "come on peeps!". I would expect a response to a question in one or two days. It's not like you were paying a tutor.

(Oh, and lose the "valley girl" vocabulary- it tends to put people off.)
 

FAQ: How to Simplify Tricky Calculus Limits?

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