I was solving a limit question that was limit(1+ax)^(1/x) as x tends to 0. I got (1)^(1/0). Please how can i continue
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2 Answers

3+3x3-3+3=
3+3x3-3+3=
Best answer

(1+ax)^(1/x)=e^a as x→0. Basis of continuous compound interest:

(1+r/n)ⁿ as n→∞.

by Top Rated User (1.0m points)

lim x-> 0 (1+ax)^(1/x)

lim x-> 0 e^ { ln (1+ax)^ (1/x)

lim x-> 0 e^ {1/x ln (1 +ax) }

lim x-> 0 e^ { ln(1 +ax)/x }

Using limit properties,

e^ { lim x-> 0 ln(1+ax)/x }

Applying L'hospitals rule we get:

e^ [ lim x-> 0 d{ln(1+ax)}/dx / d {x}/dx ]

e^ { lim x-> 0  a/(1+ax) }

e^ { a/(1+0)}

e^a

 

 

by Level 8 User (30.1k points)

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