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evaluate the limit.
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\[\lim_{x \rightarrow \infty} e^{x -x^{2}}\]
evaluate the exponent when x tends to infinity...since that tends to -infinity....so the limit will tend to e^(-infinity) which is zero
e^x*(1-x) e^infinity *(-infinity)=e^(-infinity)=1/e^infinity=0
as infinty * infinity= infinity
@mebs
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Here is what I tried. let u = -x so as x approaches -infinity u approaches infinity ...than I wrote It in terms of u
\[\huge \lim{_{u \rightarrow \infty}} e^{-u-u^{2}}\]
yeah infinity is very very large its square will also infinity
u did correctly
so its just 0 then
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