Find Limit: <Undefined,Undefined,6/5>

[Whatupdoc]

Find the limit:

lim, t->0 &lt; \frac{e^{-5t} - 1}{t}, \frac{t^{13}}{t^{14}-t^{13}}, \frac{6}{5+t}&gt;

answer: <__,__,__>

well, what i did is just plug in zero for t which i get <0,0, 6/5> which is incorrect. am i missing something? or actually it should be <undefined,undefined, 6/5>

 

[Brinx]

For the first term, try l'Hopital's rule. For the second term, see if you can write the expression using only t^{13} and lower powers of t (splitting the denominator into two multiplicative terms would help a lot), and the third term of course doesn't pose a problem.

 

[HallsofIvy]

Find the limit:

lim, t->0 &lt; \frac{e^{-5t} - 1}{t}, \frac{t^{13}}{t^{14}-t^{13}}, \frac{6}{5+t}&gt;

answer: <__,__,__>

well, what i did is just plug in zero for t which i get <0,0, 6/5> which is incorrect. am i missing something? or actually it should be <undefined,undefined, 6/5>

Well, first, as I think you understand, "plugging" in 0 does not give 0 for the first two! Neither does it give "undefined"- the limit may exist even if the value does not.

Some texts make a distinction between "a/0" when a is not 0 and "0/0". Of course, neither is a number but we often refer to "0/0" as "undetermined" rather than "undefined". "a/0" is undefined because if we try to set x= a/0 we get x*0= a which is not true for any x. If set x= 0/0, however, we get x*0= 0 which is true for all x. We still can't give a specific value for x so it is "undetermined".

This is important here because: if f(x)-> a, a nonzero number, and g(x)->0, then f(x)/g(x) must get larger and larger- there is no limit, the limit is "undefined".

If f(x)->0 and g(x)->0 also, then f(x)/g(x) may have a limit. For an obvious example, take f(x)= x and g(x)= x. As x-> 0, both of those go to 0 but their quotient is x/x= 1 (as long as x is not 0) which has limit 1 as x goes to 0.

For your example, both the first two components become "0/0" (all interesting limits do!) so you need to look more closely. You could, as Brinx suggested, use L'Hopital's rule but that is not necessary.

The second component is a little simpler than the first:
\frac{t^{13}}{t^{14}-t^{13}}
That is one polynomial divided by another. The fact that t=0 make both of them 0 means that we can factor t out of both! In fact, t¹⁴-t¹³= t¹³(t- 1). Now, you can cancel and be left with \frac{1}{t-1}. What is the limit of that as t goes to 0?

The first is a little subtler.
Do you know that the derivative of e^-5t is -5 e^-5t and so the derivative at t=0 is -5?
Do you recognize that first component as being exactly the "difference quotient" you would use in the basic definition (replacing t with "h" perhaps) to find the derivative of e^-5t at t= 0?

 

[Whatupdoc]

ahhh, i forgot all about the L'hopital rule. thanks a lot for the long explanation, i totally get it now