Could you please calculate this limit : `lim_(x->0) (2^(2x) + 2*2^x - 3)/(2^x - 1)`

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The limit `lim_(x->0) (2^(2x) + 2*2^x - 3)/(2^x - 1)` has to be determined.

`lim_(x->0) (2^(2x) + 2*2^x - 3)/(2^x - 1)`

`= lim_(x->0) (2^(2x) + 3*2^x - 2^x - 3)/(2^x - 1)`

`= lim_(x->0) (2^x(2^x + 3) - 1(2^x + 3))/(2^x - 1)`

`= lim_(x->0) ((2^x - 1)(2^x + 3))/(2^x - 1)`

`= lim_(x->0) (2^x + 3)`

Substitute x = 0

=> 1 + 3 = 4

The limit `lim_(x->0) (2^(2x) + 2*2^x - 3)/(2^x - 1) = 4`

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tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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The limit `lim_(x->0) (2^(2x) + 2*2^x - 3)/(2^x - 1)` is required.

Substituting x = 0 in `(2^(2x) + 2*2^x - 3)/(2^x - 1)` gives:

`(2^(0) + 2*2^0 - 3)/(2^0 - 1)`

= `(1 + 2 - 3)/(1 - 1)`

= `0/0`

This is an indeterminate form and we can use l'Hospital's rule to determine the limit. Substitute the numerator and denominator with their derivatives.

`(2^(2x) + 2*2^x - 3)' = ln(2)*2^(2*x+1)+ln(2)*2^(x+1)`

`(2^x - 1)' = ln(2)*2^x`

The given limit can be written as:

`lim_(x->0)(ln(2)*2^(2*x+1)+ln(2)*2^(x+1))/(ln(2)*2^x)`

= `lim_(x->0)(2^(2*x+1)+2^(x+1))/(2^x)`

Substituting x = 0 gives:

`(2^(0+1)+2^(0+1))/(2^0)`

= `(2 + 2)/1`

= 4

The required limit `lim_(x->0) (2^(2x) + 2*2^x - 3)/(2^x - 1) = 4`

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