Let `f(x)=(log_e(x^2+e^x))/(log_e(x^4+e^2x))`. If `lim_(xrarr oo) f(x)=l` and `lim_(xrarr-oo)f(x)=m`, then

Options

A. `l=m`
B. `l=2m`
C. `2l=m`
D. `l+m=0`

Answer

A

Explanation

We have,
`l=lim_(xtooo)f(x)=lim_(xtooo)(log_e(x^2+e^x))/(log_e(x^4+e^(2x)))=lim_(xtooo)(2x+e^x)/(x^2+e^x)xx(x^4+e^(2x))/(4x^3+2e^(2 x))`
` rArr l=lim_(xtooo) ((2(x)/(e^x)+1)((x^4)/e^(2x)+1))/((1+(x^2)/(e^x))(2+(4x^3)/e^(2x)))`
` rArr l=((2xx0+1)(0+1))/((1+0)(+0))=(1)/(2)`
`rArr m=lim_(xto-oo)((2+(e^x)/(x))(1+(e^(2x))/(x^4)))/((4+(2e^(2x))/(x^3))(1+(e^x)/x^2))=((2+0)(1+0))/((4+0)(1+0))=(1)/(2)`
`therefore m=l`.