Need help with limits

1) I did one mistake here, I'll answer the 1st question right away.



2) the function h(x)=|x|/x is continuous at 3, so again we have: limx->3h(x)=h(3)=I3I/3=1.


We only would have problem if we were asked to find the limit when x approaches 0, because if x<0 then h(x)=-1 and if x>0 h(x)=1. So the limit x->1 does not exist at this function.

Post Merge: 11 years ago

1) To find the limit as x approaches 3, we must firstly find the limit as x approaches 3- (from the negative side, x<3) and then find the limit as x approaches 3+ (from the positive side, x>3). Then we compare the limits: if they are the same number, then the limit exists and it is this same number. Otherwise, if the 3- and 3+ limits are different, then the limit as x aproaches to 3 does not exists.

So, when x<3 we have: f(x)=lnx. So  limx->3-f(x)= limx->3-lnx=ln3 (cause lnx is continuous at 3)
and, when x>3 we have: f(x)=x2*ln3 So limx->3+f(x)= limx->3+x2*ln3=32*ln3= 9ln3 (cause x2*ln3 is continuous at 3)

But since limx->3-f(x) is different from limx->3+f(x), (because ln3 is different from 9ln3) the limit limx->3f(x) does not exist.

Thanks so much for the help, this was practice for my exam!