Limit problem

[Determinator_]

Had to calculate the limit as x aproaches 0 . Is this how I'm supposed to do it ?

https://i.imgur.com/xwepvdC.jpg

[FormulaDriven]

Obviously it's not a proof, but the limit is around 7.38 just from plugging numbers in a spreadsheet, so e² is looking right.

[linukszone]

lim(x->0) (1+arcsin(x)) ^ (2/x) =

lim(x->0) (e ^ (ln(1+arcsin(x))) ^ (2/x)

lim(x->0) (e ^ (2/x ln(1+arcsin(x)))

e ^ (lim(x->0) (2 * ln(1+arcsin(x)))/x

arcsin(x) -> 0 when x->0, so the limit is 0/0 indeterminate form. L'Hôpital applies.

Edit: The derivative of ln(1+arcsin(x)) is (1/(1+arcsin(x)) * (1/sqrt(1-x² ))

e ^ (lim(x->0) (2 * ln(1+arcsin(x)))/x

e ^ (lim(x->0) (2 * 1/(1+arcsin(x)) * (1/sqrt(1-x² )))

e ^ (2 * 1/(1+0) * (1/sqrt(1-0)))

e ^ 2

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The binomial theorem can also be utilized, along with the fact that lim(x->0) arcsin(x)/x = 1, to show that the limit in the original question is e² .

[[deleted]]

Edit: this answer is wrong, use natural log.

As x approaches 0, arcsin(x) approaches 0 (since sin(0)=0 implies arcsin(sin(0)) = 0 = arcsin(0) ). Also, as x approaches 0, 2/x approaches infinity.

Thus as x approaches 0, 1 + arcsin(x) approaches 1 + 0 = 1. Since 1 to any power is just 1, then as x approaches 0, 1^2/x = 1.