Find`lim_(t->0) (sin(2t)/tveci+(t-2)^5vecj+tlntveck)`

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txmedteach | High School Teacher | (Level 3) Associate Educator

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Because of the orthogonal vectors, `veci,vecj, and veck`, we can consider the limits of each function of t separately.

Let's consider the first limit.
`lim_(t->0) sin(2t)/t`

Clearly, if we were to evaluate the limit from the left or right, we get a result of `0/0`. This, believe it or not, is a very fortunate result, because we can now use L' Hopital's rule:

`lim_(t->0) sin(2t)/t = lim_(t->0) (d/dtsin(2t))/(d/dtt)`

Let's simplify the derivatives:

`lim_(t->0) (2cos(2t))/1 = 2`

So, the first limit is 2.

Let's move on to the second limit:

`lim_(t->0) (t-2)^5`

We can evaluate this limit explicitly by letting t = 0. This limit's value is:

`lim_(t->0) (0-2)^5 = -32`

Finally, we can find the third limit:

We'll need to assume that t can only approach 0 from the right (in other words, `tepsi[0,oo]`). Otherwise, the third limit does not exist, and the whole problem breaks down. Of course, if t is not restricted in this way, then the answer to the problem is that the limit does not exist!

`lim_(t->0^+)tlnt = lim_(t->0^+) lnt/(1/t)`

We have come up, again, on another L' Hopital situation:

`lim_(t->0^+) lnt/(1/t) = lim_(t->0^+) (1/t)/(-1/t^2)`

We can simplify:

`=lim_(t->0^+) -t = 0`

So, if our restrictions on t hold, then the third limit is 0.

Therefore, our full limit is `2veci-32vecj`. Again, this limit holds as long as t is only defined from 0 to infinity. If the domain of t is `RR` then our limit does not exist because the third limit from the left does not exist.


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