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lim x tends to infinity [ln (x^2 + e^x)]/ln[(x^4 + e^(2x))] ?
Limx-infinity [ln (x2+ ex)]/[ln ( x4+ex)] So expanding ex and e2x we get; Linx-infinity [ln (x2+1+x+x+x2/2!+.........)]/[ln(x4+1+2x+(2x)2/2!........)] So taking xn 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
Limx-infinity [ln (x2+ ex)]/[ln ( x4+ex)]
So expanding ex and e2x we get;
Linx-infinity [ln (x2+1+x+x+x2/2!+.........)]/[ln(x4+1+2x+(2x)2/2!........)]
So taking xn 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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