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How do you find the limit #lima^x# as #x->oo#?

1 Answer

Dec 31, 2016

Use #a^x = e^(xlna)# and consider cases.

Explanation:

For #0 < a < 1#, we have #lna < 0#, so

as #xrarroo#, we get #xlnararr-oo# and so #a^x rarr 0#.

For #a=0#, we get the indeterminate form #1^oo#.

For #a > 1#, we have #lna > 0#, so

as #xrarroo#, we get #xlnararroo# and so #a^x rarr oo#.

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