Question

what is the limit of (2-(e)^x/2)^4/x as x approaches 0??

Answer 1

is it infinite?

Answer 2

\[\lim_{x \rightarrow 0} \ \ (2-e^{\frac{x}{2}})^{\frac{4}{x}}\]

Answer 3

yup

Answer 4

letting x=0 it becomes \[1^\infty\]undefined

Answer 5

?

Answer 6

if I apply the LHR?

Answer 7

yeah LHR

so let\[y= (2-e^{\frac{x}{2}})^{\frac{4}{x}}\]\[\ln y=\frac{4}{x} \ln(2-e^{\frac{x}{2}})=4\frac{\ln(2-e^{\frac{x}{2}})}{x}\]apply the LHR for right hand side

Answer 8

that would be?

Answer 9

can u apply LHR for\[\frac{\ln(2-e^{\frac{x}{2}})}{x}\]as x--->0

Answer 10

\[\frac{ \frac{ x }{ 2 }(-e ^{\frac{ x }{ 2}}) }{ 2-e ^{\frac{ x }{ 2 }}}\]?

Answer 11

well for 1^(infinity) forms,,remember this for convenience ..
ans would be e^(f(x).g(x))
where we had something like this :
(1+f(x))^g(x) ,,here f(x)-->0 and g(x)-->infinity..

Answer 12

in your case,
f(x) = 1- e^(x/2)
and g(x) = 4/x

ans would be:
e^[ (x-->0) (1-e^(x/2) * 4/x ]

you may apply LH rule here,,

Answer 13

was i able to explain well ?

Answer 14

still don't get it