# problems on limits of sequences

• Apr 8th 2007, 07:20 PM
luckyc1423
problems on limits of sequences
1)
A) Lim x->0+ (1/x)
B) Lim x->0+ |sin(1/x)|
C) Lim x->0+ xsin(1/x)

2)
Determine the following limits:
A) Lim x->1 (x^3+5)/(x^2+2)
B) Lim x->1 (sqrt(x)-1)/(x-1)
C)Lim x->0 (x^2 + 4x)/(x^2+2x)
D) Lim x->0 (sqrt(4+x) - 2)/x
E) lim x->0- 4x/|x|
• Apr 8th 2007, 07:50 PM
qbkr21
Quote:

Originally Posted by luckyc1423
1)
A) Lim x->0+ (1/x)
B) Lim x->0+ |sin(1/x)|
C) Lim x->0+ xsin(1/x)

2)
Determine the following limits:
A) Lim x->1 (x^3+5)/(x^2+2)
B) Lim x->1 (sqrt(x)-1)/(x-1)
C)Lim x->0 (x^2 + 4x)/(x^2+2x)
D) Lim x->0 (sqrt(4+x) - 2)/x
E) lim x->0- 4x/|x|

I am having somewhat of a difficult time understanding your notation for the first section of problems. What are you trying to indicate by using the plus (+) sign? Also for problems D which part is under the square root? sqrt(4+x)-2? Why not write 2-x? Some clarification would be sweet.

http://item.slide.com/r/1/22/i/APb-a...rG9KQCkvTiVJM/
• Apr 8th 2007, 07:59 PM
luckyc1423
The 0+ I think means you are approaching 0 from the positive direction

the 4+x is the only thing in the sqrt

sqrt(4+x)
• Apr 8th 2007, 08:00 PM
luckyc1423
so its sqrt(4+x) then minus 2 and all of that divided by x
• Apr 8th 2007, 08:01 PM
qbkr21
Re:
Here is a Graph for #2 E. This is one of those indeterminate problems there for you must take the limit from both the right and left hand side. It is easy to see by looking the graph. The left-hand limit is -4, and just as you can tell the right hand limit is 4. Why? Because the function is zeroing in on the Y axis to -4 and 4 respectfully. http://item.slide.com/r/1/56/i/aAuwl...5tp-QZHoZM1Pt/
• Apr 8th 2007, 08:07 PM
qbkr21
Re:
Quote:

Originally Posted by luckyc1423
so its sqrt(4+x) then minus 2 and all of that divided by x

Ohh...In this case it now big deal you just have to multiply by a factor of one. Any time you initial solution yields indeterminate 0/0 or inf/inf or -inf/-inf or -inf/inf or inf/-inf you need to factor or use algebra.

http://item.slide.com/r/1/17/i/OOszf...bMJPt92GjdTEB/
• Apr 8th 2007, 08:39 PM
Jhevon
Quote:

Originally Posted by luckyc1423
1)
A) Lim x->0+ (1/x)

notice as we approach 1/x from the right, we have positive values. so our limit will be positive.

now lim{x--> 0+} 1/x = +oo, since 1/x --> oo as x --> 0 and we have a positive limit

Quote:

B) Lim x->0+ |sin(1/x)|
we saw above that as x-->0+, 1/x --> +oo

so for lim{x-->0+} |sin(1/x)|, we have sin(1/x)-->sin(+oo) as x-->0+, now sine oscillates between 1 and -1 for it's maximum and minimum values. since we have |sin(1/x)| however, all negative values become positive. however, we still have oscillations as we change x. so the limit does not exist

Quote:

C) Lim x->0+ xsin(1/x)
lim{x-->0+}xsin(1/x) = lim{x-->0+}x * lim{x-->0+}sin(1/x) = 0*lim{x-->0+}sin(1/x) = 0
• Apr 8th 2007, 08:41 PM
qbkr21
Re:
Ohh...Yea thanks Jhevon. Since Latex is down I couldn't tell why the plus (+) sign was being used. Duhh... Approach from the right. Hopefully they are still working on getting it fixed. Thanks!