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lim(xto0)(a^(sinx)-1)/(b^(sinx)-1) is eq...

`lim_(xto0)(a^(sinx)-1)/(b^(sinx)-1)` is equal to

A

`a/b`

B

`b/a`

C

`(log_(e)a)/(log_(e)b)`

D

`(log_(e)b)/(log_(e)a)`

Text Solution

Verified by Experts

The correct Answer is:
C
`lim_(xto0)(a^(sinx)-1)/(b^(sinx)-1)`
Using L'Hospital 's rule
`=lim_(xto0)(log_(e)a.a^(sinx))/(log_(e)b.b^(sinx))=(log_(e)a)/(log_(e)b)`
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