Math problem, please solve..
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Math problem, please solve..

[From: ] [author: ] [Date: 12-01-23] [Hit: ]
Differentiate numerator and denomenator.Log 4 cancels and also 4^x.So the answer will be 1/4.And since 1/(4^x) goes to zero as x -> +inf,use Daiiis method. no need to use l hospitals rule here.......
The limit of this as x approaches positive infinity:

( ( 4^(x) ) - 1 ) / ( 4^(x+1) )

Thanks.

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Use L'Hospital's rule.

Differentiate numerator and denomenator.

You will get 4^xlog4/4^(x+1)log4
Log 4 cancels and also 4^x.

So the answer will be 1/4.

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( ( 4^(x) ) - 1 ) / ( 4^(x+1) )

= (1/4) [ 1 - 1/(4^x) ]

And since 1/(4^x) goes to zero as x -> +inf, the original expression goes to 1/4

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1/4

use Daiii's method. no need to use l' hospitals rule here.
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