# Calculate limit with special limits limit (2^x+3^x-2)/(x^2+x) x-->0

Asked on by minlux

sciencesolve | Teacher | (Level 3) Educator Emeritus

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Since the expression of function whose limit you need to evaluate contains exponentials, you may use the following special limit, such that:

`lim_(x->0)(b^x - 1)/x = ln b`

You may re-write the numerator of the function such that:

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

You may group the terms, such that:

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

You may factor out x to denominator, such that:

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

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

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

Using the property of limit, yields:

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

Using the formul of special limit, yields:

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

Converting the sum of logarithms into the logarithm of product, yields:

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

`lim_(x->0) (2^x+ 3^x - 2)/(x^2+x) = ln 6`

Hence, evaluating the limit, using the formula of special limit `lim_(x->0)(b^x - 1)/x = ln b` , yields `lim_(x->0) (2^x+ 3^x - 2)/(x^2+x) = ln 6.`

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