lim x tends to infinity [ln (x^2 + e^x)]/ln[(x^4 + e^(2x))] ?

Question

chandradeep pokhariya · Answer

Lim x-infinity [ln (x 2 + e x )]/[ln ( x 4 +e x )] So expanding e x and e 2x we get; Lin x-infinity [ln (x 2 + 1+x+x+x 2 /2!+.........)]/[ln(x 4 +1+2x+(2x) 2 /2!........)] So taking x n common in numerator and denominator we can apply limit and ln (x^n)/ln(x^n) will be obtained so they can be cancelled and the answer will be 1

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lim x tends to infinity [ln (x^2 + e^x)]/ln[(x^4 + e^(2x))] ?

lim x tends to infinity [ln (x^2 + e^x)]/ln[(x^4 + e^(2x))] ?

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