# calculate lim ln(x^2+e^5x)/ln(x^10+e^4x)  x-->0

You need to substitute 0 for x such that:

`lim_(x->0) (ln(x^2+e^(5x)))/(ln(x^10+e^(4x))) = ln(0^2+e^0)/ln(0^10+e^0)`

`lim_(x->0) (ln(x^2+e^(5x)))/(ln(x^10+e^(4x))) = (ln(0+1))/(ln(0+1))`

`lim_(x->0) (ln(x^2+e^(5x)))/(ln(x^10+e^(4x))) = ln 1/ln 1`

`lim_(x->0) (ln(x^2+e^(5x)))/(ln(x^10+e^(4x))) = 0/0`

Since the limit is indeterminate, `0/0` , you may use l'Hospital's theorem such that:

`lim_(x->0) (ln(x^2+e^(5x)))/(ln(x^10+e^(4x))) = lim_(x->0) ((ln(x^2+e^(5x)))')/((ln(x^10+e^(4x)))') `

`lim_(x->0) (ln(x^2+e^(5x)))/(ln(x^10+e^(4x))) =...

You need to substitute 0 for x such that:

`lim_(x->0) (ln(x^2+e^(5x)))/(ln(x^10+e^(4x))) = ln(0^2+e^0)/ln(0^10+e^0)`

`lim_(x->0) (ln(x^2+e^(5x)))/(ln(x^10+e^(4x))) = (ln(0+1))/(ln(0+1))`

`lim_(x->0) (ln(x^2+e^(5x)))/(ln(x^10+e^(4x))) = ln 1/ln 1`

`lim_(x->0) (ln(x^2+e^(5x)))/(ln(x^10+e^(4x))) = 0/0`

Since the limit is indeterminate, `0/0` , you may use l'Hospital's theorem such that:

`lim_(x->0) (ln(x^2+e^(5x)))/(ln(x^10+e^(4x))) = lim_(x->0) ((ln(x^2+e^(5x)))')/((ln(x^10+e^(4x)))') `

`lim_(x->0) (ln(x^2+e^(5x)))/(ln(x^10+e^(4x))) = lim_(x->0) ((x^2+e^(5x))')/(x^2+e^(5x))*(x^10+e^(4x))/((x^10+e^(4x))')`

`lim_(x->0) (ln(x^2+e^(5x)))/(ln(x^10+e^(4x))) = lim_(x->0) (2x + 5e^(5x))/(x^2+e^(5x))*(x^10+e^(4x))/(10x^9 + 4e^(4x))`

Substituting 0 for x yields:

`lim_(x->0) (ln(x^2+e^(5x)))/(ln(x^10+e^(4x))) = (0+5e^0)/(0+e^0)*(0+e^0)/(0+4e^0)`

`lim_(x->0) (ln(x^2+e^(5x)))/(ln(x^10+e^(4x))) = 5/1*1/4`

`lim_(x->0) (ln(x^2+e^(5x)))/(ln(x^10+e^(4x))) = 5/4`

Hence, evaluating the given limit yields `lim_(x->0) (ln(x^2+e^(5x)))/(ln(x^10+e^(4x))) = 5/4.`

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