Question 1165554
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lim (x-> -inf) (x^2+4^-x)/(x^2-4^x)
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<pre>
To answer this question, you should apply your knowledge of Calculus.


The Calculus says that in the numerator, when  x --> {{{-infinity}}},  the term {{{4^(-x)}}}  dominates 
and goes to infinity much fster than {{{x^2}}} does.


In the denominator,  when  x --> {{{-infinity}}},  the term  {{{x^2}}}  dominates comparing with  {{{4^x}}}.


Therefore, in whole, as x goes to  {{{-infinity}}},  the given function behaves as  {{{4^(-x)/x^2}}}.


When  x --> {{{-infinity}}},  the numerator of this reduced fraction goes to infinity much faster than the denominator.


It gives the <U>ANSWER</U>:  the given function goes to {{{infinity}}}  as  x --> {{{-infinity}}}.
</pre>

Solved.