# Hello! Have a lim x->0 ( ln (3x+1) / (2x^2)+7 ) I think, may be lim x->0 ( ln (3x+1)/x^2 / ((2x^2)+7) /x^2 ) but ln crashed.

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### 1 Answer

Lt x--> 0 (ln(3x+1)/(2x^2)+7)

Here the way the brakets are put the 7 is not in denominator. Actually 7 is a separate term and the given limit is rewitten as follows:

Lt x--> 0 {ln(3x+1)/2x^2 + 7 .

Then Lt x--> 0 ln(3x+1)/2x^2 + 7

But lt x--> 0 ln(3x+1)/2x^2 is 0/0 form indeterminate.

So we go for L'Hospital's rule of defferentiating numerator and denominator and then taking the limit.

Therfore (ln(3x+1)'/(2x^2)' = 3/(3x+1)(4x).

Therefore Ltx--> 0 ln(3x+1)/(2x^2) = Lt 3/(3x+1)(4x) = 1/(3*0+1)(4*0) = infinity.

2)

Lt x--> 0 (ln(3x+1)/x^2 divided by ((2x^2+7)/x^2) . Here the 7 is taken inside the braket. Thus this is different expression.

Therefore lt x--> 0 (ln(3x+1))/x^2 / (2x^2+7)/x^2) = Ltx--> 0 ln(3x+1)/(2x^2+7) = 0/ (0+7) = 0/7 = 0 and this not an indeterminate form . So we can get the limit by just substituting x= 0 in the expression.