# What is limit x tends to infinity (x^2 - e^x) / (1 + e^x)

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### 1 Answer

We are required to find the value of : lim x--> inf. [(x^2 - e^x) / (1 + e^x)]

Substituting x = inf., gives the indeterminate form 0/inf. But l'Hopital's rule can be used only for the forms 0/0 and inf./inf. So we make some conversions.

lim x--> inf. [(x^2 - e^x) / (1 + e^x)]

=>lim x--> inf. [(x^2/(1 + e^x)] - lim x--> inf. [(e^x)/(1 + e^x)]

substituting x = inf. here gives inf./inf. in both the cases, so we can replace the numerators and denominators by their derivatives.

lim x--> inf. [(2x/(e^x)] - lim x--> inf. [(e^x)/(e^x)]

lim x--> inf. [(2x/(e^x)] - lim x--> inf. [1]

Again substituting x = inf. the first limit is inf./inf., and the second is 1. Use l'Hopital's rule for the first.

lim x--> inf. [(2/(e^x)] - 1

Substituting x = inf. gives us 0 for the first limit as 1/inf. = 0 and the second is 1.

The final result is -1.

**The value of : lim x--> inf. [(x^2 - e^x) / (1 + e^x)] = -1.**