Question

whats the limit t-->0^+ of (9t^t)?

Answer 1

is this
\[\lim_{t\rightarrow 0^+} 9t^t\]?

Answer 2

can take the 9 right outside because it is a constant and compute
\[9\lim_{t\rightarrow 0^+} t^t\]

Answer 3

to compute this limit, you have a choice. either "take the log" or rewrite
\[t^t=e^{t\ln(t)}\] but all the work is the same
you were probably told to "take the log"

Answer 4

yeah I know it's l'hopitals rule but I can't get it o work

Answer 5

yeah take the log

Answer 6

ok so first step is take the log and get
\[t\ln(t)\] right?

Answer 7

then it is of the form
\[0\times -\infty\] so you use the standard trick of rewriting it as
\[\frac{\ln(t)}{\frac{1}{t}}\] and now it looks like
\[\frac{\infty}{\infty}\] so you can use l'hopital

Answer 8

the derivative of the log is
\[\frac{1}{t}\] and the derivative of
\[\frac{1}{t}\] is
\[-\frac{1}{t^2}\]

Answer 9

but if you put in 0 it's still infinity?

Answer 10

so after l'hopital you get
\[\frac{\frac{1}{t}}{-\frac{1}{t^2}}=-t\]

Answer 11

now the limit is clearly 0, so your original limit is
\[9e^0=9\]

Answer 12

Oh I see that make sense thanks a lot for your help

Answer 13

clear that we have to "exponentiate" right because the first step was to "take the log" so we computed the limit of the log, not the limit we were after

Answer 14

yw, and this is a pretty common limit, so you might just remember that
\[\lim_{x\rightarrow 0^+}x^x=1\]