`f(x)=("ln"(x^(2)+e^(x)))/("ln"(x^(4)+e^(2x)))`. Then `lim_(x to oo)` f(x) is equal to

Options

A. `(2a)/(pi)`
B. `-(2a)/(pi)`
C. `(4a)/(pi)`
D. `-(4a)/(pi)`

Answer

B

Explanation

`L=underset(xtooo)lim("ln"(x^(2)+e^(x)))/("ln"(x^(4)+e^(2x)))=underset(xtooo)lim("ln "e^(x)(1+(x^(2))/(e^(x))))/("ln "e^(2x)(1+(x^(4))/(e^(2x))))`
`=underset(xtooo)lim(x+"ln "(1+(x^(4))/(e^(x))))/(2x+"ln "(1+(x^(4))/(e^(2x))))`
`=underset(xtooo)lim(1+(1)/(x)" ln "(1+(x^(2))/(e^(x))))/(2+(1)/(x)" ln "(1+(x^(4))/(e^(2x))))`
Note that as `xtooo,(x^(2))/(e^(x))to0` and as `xtooo,(x^(2))/(e^(2x))to0`
(Using L'Hospital's rule)
Hence `L=(1)/(2)`.