
#1
Oct206, 12:48 AM

P: 26

[tex]\lim_{x\rightarrow\infty}\frac{\sinx}{x}[/tex]
[tex]\lim_{x\rightarrow 0}\frac{xx2}{x1}[/tex] thanks! 



#2
Oct206, 01:32 AM

HW Helper
P: 3,353

ok hello people this is my first post! woot! okily i gtg to da shops now, so basically i just answer ur post bout da limits...ok well the first one, sin lxl divided by x, approaching infinity, well you see, think about it this way, sin lxl has to be between 1, and 1, as we know. now, 1/infinty= 0
1/infinity=0=0, anything inbetween will also equal zero, except zero it self, but as u can see, the limit as it approaches that is also zero, i know im not very clear..ok basically its zero..and the 2nd one, a little easier :D well basically people use limits to find out wat the value something wud approach at an asymtote or a discontinous point. however, in this case, there is no discontinuity at point zero, so you can just sub it straight in and you'll be fine. so its (0  l02l)/(01)= (0 +2)/1= 2 i know my layout is bad, i hope u get da gist of it.. 



#3
Oct206, 04:43 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

For the first, since x is approaching infinity, x is always positive and x= x.
For the second, do the two onesided limits: For x close to 0 and x> 0, x= x and x 2= 2 x ( x is close to 0 so x< 2, x 2 is negative). For x close to 0 and x< 0, x= x and x2= 2 x (x is still less than 2). (Gib Z is correct since the second function is continuous at x= 0, you can just substitute x= 0 but it might be that you want to find the limit in order to show that the function is continuous.) 


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