Let f(x) = frac{ln(x^2 + e^x)}{ln(x^4 + e^{2x | vedclass

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(A) For $x 	o \infty$,$f(x) = \frac{\ln(e^x(1 + x^2/e^x))}{\ln(e^{2x}(1 + x^4/e^{2x}))} = \frac{x + \ln(1 + x^2/e^x)}{2x + \ln(1 + x^4/e^{2x})}$.Divi

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$l = m$

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(A) For $x 	o \infty$,$f(x) = \frac{\ln(e^x(1 + x^2/e^x))}{\ln(e^{2x}(1 + x^4/e^{2x}))} = \frac{x + \ln(1 + x^2/e^x)}{2x + \ln(1 + x^4/e^{2x})}$.Dividing numerator and denominator by $x$,we get $\lim_{x 	o \infty} f(x) = \frac{1 + 0}{2 + 0} = \frac{1}{2}$,so $l = 1/2$.For $x 	o -\infty$,let $t = -x$,so $t 	o \infty$. Then $f(-t) = \frac{\ln(t^2 + e^{-t})}{\ln(t^4 + e^{-2t})}$.As $t 	o \infty$,$e^{-t} 	o 0$ and $e^{-2t} 	o 0$.Thus,$m = \lim_{t 	o \infty} \frac{\ln(t^2)}{\ln(t^4)} = \lim_{t 	o \infty} \frac{2 \ln(t)}{4 \ln(t)} = \frac{2}{4} = \frac{1}{2}$.Since $l = 1/2$ and $m = 1/2$,we have $l = m$.

Let $f(x) = \frac{\ln(x^2 + e^x)}{\ln(x^4 + e^{2x})}$. If $\lim_{x \to \infty} f(x) = l$ and $\lim_{x \to -\infty} f(x) = m$,then:

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