would anyone be able to help me understand why the graph has a value for when x = 0. i imagine it's similar to sinx/x and the limit being zero but haven't been able to find a proof or explanation on youtube.
pic of the graph for reference
#graph of (e^x - 1)/ x
uncut nebula
merry gardenBOT
fickle veldt
would anyone be able to help me understand why the graph has a value for when x ...
$e^0 = 1$ is defined.
Suppose $a \neq 0\$
Then $a^0 = a^{1-1} = a^1 \cdot a^{-1} = a^1 \cdot \frac{1}{a^1} = \frac{a}{a} = 1\$
barren houndBOT
adonhs
fickle veldt
Ah wait I see it now..
uncut nebula
yea haha
fickle veldt
$y=\frac{e^{x}-1}{x}$
barren houndBOT
adonhs
fickle veldt
See as x approaches 0 we get 0/0
uncut nebula
yes
fickle veldt
You could use L'Hopital Bernoulli to see that the limit is 1..
uncut nebula
ooh okay thank you, i haven't covered that in school yet so i'll watch a video on it
you differentiate top and bottom?
fickle veldt
Yes it's similiar to sin(x)/x if this was the question.
fickle veldt
you differentiate top and bottom?
Yes, if you have the case infinity/infinity or 0/0
uncut nebula
so do you mean if the limit is infinity i gain nothing by doing it?
fickle veldt
so do you mean if the limit is infinity i gain nothing by doing it?
yes because according to my professor if you get infinity as a result after lhopital than that means that the limit cannot be determined with l'hopital
barren houndBOT
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fickle veldt
thanks youve been lots of help
.solved
merry gardenBOT
Solved