What is the limit of #e^(2x) / x^2# as x approaches infinity?
1 Answer
Feb 29, 2016
Explanation:
Since both the numerator and denominator approaches infinity, we can use L'Hospital Rule.
#lim_{x -> oo} frac{e^{2x}}{x^2} = lim_{x -> oo} frac{frac{"d"}{"d"x}(e^{2x})}{frac{"d"}{"d"x}(x^2)}#
#= lim_{x -> oo} frac{2e^{2x}}{2x}#
#= lim_{x -> oo} frac{e^{2x}}{x}#
The above expression is again indeterminate and we have to use L'Hospital Rule again
#lim_{x -> oo} frac{e^{2x}}{x} = lim_{x -> oo} frac{frac{"d"}{"d"x}(e^{2x})}{frac{"d"}{"d"x}(x)}#
#= lim_{x -> oo} frac{2e^{2x}}{1}#
#= +oo#
