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OpenStudy (anonymous):
The limit
lim (5^(x)-1)/x
x->0
represents the derivative of some function f(x) at some number a. Find f and a.
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OpenStudy (anonymous):
\[\lim_{x \rightarrow 0}5^{x-1} \div x?\]
OpenStudy (turingtest):
l'hospital's rule
OpenStudy (anonymous):
or\[\lim_{x \rightarrow 0}5^{x}-1 \div x\]?
OpenStudy (anonymous):
second one
OpenStudy (turingtest):
\[\lim_{x\to0}{5^x-1\over x}\]
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OpenStudy (anonymous):
yup @TuringTest
OpenStudy (turingtest):
use l'Hospital if you are allowed to; take the derivative of the top and bottom, then take the limit again
OpenStudy (anonymous):
well im not allowed to use l'hoapital rule
OpenStudy (turingtest):
hm....
OpenStudy (anonymous):
|dw:1340305376609:dw|
then u will get indeterminant of the form 0/0
you apply L'HOSPITALS RULE
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OpenStudy (turingtest):
@msamido the asker just stated that we cannot use l'Hospital here, so we must find another way
OpenStudy (anonymous):
...but i was thinking that ....it is indeterminant of the form 0/0 ...
OpenStudy (turingtest):
yeah\[{5^0-1\over0}=\frac00\]and l'Hospital \(does\) work, but apparently it's not allowed :(
OpenStudy (zarkon):
are you allowed to use one of the definitions of \(e\)
\[\lim_{x\to0}{e^x-1\over x}=1\]
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