The values 108, 3, 0, and 7/3 are the solutions to the given limit problems.

[celeste6]

hey could anyone help me with some limit probs?
the lim as x approaches 9
(x^2 - 81)/($$\sqrt{}x$$ - 3)

lim as x approaches 0
(2x+sinx)/x

lim as x approaches o from the right side
(sinx)/(5$$\sqrt{}x$$)

lim as x approaches 0
(tan7x)/(sin3x)

i was thinking 0 for the 1st 2 and und for the last two but I'm not sure
thanks!

 

[Dick]

All four of your guesses would be wrong. Maybe it's time to stop guessing.

 

[rock.freak667]

Here is a hint...if you get the indeterminate form of 0/0 and the function is all in x then something must be factored and canceled.
and for the rest,try to get everything in terms of limits that you know for example; you know that $$lim_{\rightarrow{0}} \frac{sin(x)}{x} =1$$ then try to reduce everything to become some function* sinx/x

 

[dynamicsolo]

hey could anyone help me with some limit probs?
the lim as x approaches 9
(x^2 - 81)/($$\sqrt{}x$$ - 3)

lim as x approaches 0
(2x+sinx)/x

lim as x approaches o from the right side
(sinx)/(5$$\sqrt{}x$$)

lim as x approaches 0
(tan7x)/(sin3x)

It's always a good idea to pay attention to the one-sided limits when the function has an obvious vertical asymptote. This can be an issue on the problems that ask for two-sided limits.

By the way, I believe each of these problems calls for a different technique...

 

[CompuChip]

• $$\lim_{x\to9} (x^2 - 81)/(\sqrt{x} - 3)$$
  This is of the form 0/0. Do you know L'Hôpitals rule? Otherwise try Taylor expansion.
  
• $$\lim_{x \to 0} (2x+\sin x)/x$$
  
  Try writing this as $\lim_{x \to 0} f(x) + g(x)$ and taking the limits separately.
  
• $$\lim_{x\downarrow0} (\sin x)/(5 \sqrt{x})$$
  
  This is again of the form 0/0. Note that you don't have to do anything special with the "from the right side" phrase, it is there only to make sure that the square root exists during the limit process.
  
• $$\lim_{x\to0} (tan7x)/(sin3x)$$
  Write this as $$\frac{\sin 7x}{\sin 3x} \cdot \cos 7x$$.
  The cosine tends to 1, so you need only worry about the quotient of the sines. Try to apply what you've learned to it.

 

[Gib Z]

Maybe its time to stopping being an ass.

My god, chill out dude, dick's just telling the original poster to start working instead of just guessing. I think we all just need to think a little more clearly, no special rules or taylor series are required here!

[itex]a^2-b^2=(a+b)(a-b)[/tex]
Noting that $x^2-81 = (\sqrt{x})^4 - 3^4$, apply difference of two squares twice, cancel out common factors and sub in x=9, easy.

The second one, do it just like CompuChip, and use $lim_{x\to 0} \frac{\sin x}{x} =1$. Remember this limit by the way, its useful.

For the third one, why not rewrite $$\frac{\sin x}{5\sqrt{x}}$$ as $$\frac{\sqrt{x}}{5} \cdot \frac{\sin x}{x}$$ =D?

For the last one, as CompuChip said, the cosine tends to 1. For the sines, think about what the sinx/x limit means, and apply it.

 

[VietDao29]

Maybe its time to stopping being an ass.

And now, it' time to stop being rude like that! We're not paid to help you. Think about it, did you try anything? Why must we help you? After all, it your homework, not ours.

Guessing, eh? Base on what reason?? Your sixth sense, huh? I could guess the final result for hundreds of problems in less than a blink. Be more co-operative, please. We are here to help, not to provide solutions.

 

[dynamicsolo]

And now, it' time to stop being rude like that! We're not paid to help you. Think about it, did you try anything? Why must we help you? After all, it your homework, not ours.

Guessing, eh? Base on what reason?? Your sixth sense, huh? I could guess the final result for hundreds of problems in less than a blink. Be more co-operative, please. We are here to help, not to provide solutions.

Erm, please note that the rude individual was not the person who posted the problems (who has not yet responded).

As a counterpoise, many of the posters have been a little too nice in providing methods without asking the OP to do any thinking of their own...

 

[HallsofIvy]

Actually, my first thought was that gomerpyle was telling celeste6 not to be an ass!
Which is still rude, of course.

 

[celeste6]

o, I'm sorry, I'm new to this site. thanks for your help though,
so they would be
108
3
0
7/3
:)

 

[dynamicsolo]

o, I'm sorry, I'm new to this site. thanks for your help though,
so they would be
108
3
0
7/3
:)

I believe we would agree with those answers. Glad we could help.

 

[HallsofIvy]

o, I'm sorry, I'm new to this site. thanks for your help though,
so they would be
108
3
0
7/3
:)

If I were your teacher I would insist upon your writing each answer as a complete sentence! WHAT is "108", 3, 0, 7/3?

 

[celeste6]

If I were your teacher I would insist upon your writing each answer as a complete sentence! WHAT is "108", 3, 0, 7/3?

Those are the answers, I'm sorry I didn't write in full sentences. I have MD and didn't have my microphone at the time.