How do you find the limit of #e^(-5x)# as x approaches #oo#?

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Answer 1 · 2016-04-13T06:11:53

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#e^(ax)=1+(ax)+(ax)^2/2!+(ax)^3/3!+...to oo#, as #xtooo#, when a > 0.

So, th reciprocal #e^(-ax) to 0#, as #x to oo#.

Here, a = 5 > 0.

Answer 2 · 2016-04-13T11:14:48

The limit is #0#. Here is an informal approach.

As #x# increases without bound (as #x# gets bigger and bigger),

the product #5x# also gets bigger and bigger,

so, the exponent #-5x# is a negative number farther and farther from #0# (more and more negative) ( big negative number)

Wait, #e^(-5x) = 1/e^(5x)#.

Now #x rarroo#, so #5x rarroo# and #e^(5x) rarroo#.

So we have #1# over a number that is going to infinity, so the ratio is going to #0#.