How do you evaluate #[ ( 1 + 3x )^(1/x) ]# as x approaches infinity?

1 Answer
Euan S.
Aug 13, 2016

#lim_(xrarroo)(1+3x)^(1/x) = 1#

Explanation:

Going to use a nifty wee trick that makes use of the fact that the exponential and natural log functions are inverse operations. This means we can apply both of them without changing the function.

#lim_(xrarroo)(1+3x)^(1/x) = lim_(xrarroo)e^(ln(1+3x)^(1/x))#

Using the exponent rule of logs we can bring the power down in front giving:

#lim_(xrarroo)e^(1/xln(1+3x))#

The exponential function is continuous so can write this as

#e^(lim_(xrarroo)1/xln(1+3x))#

and now just deal with the limit and remember to sub it back into the exponential.

#lim_(xrarroo)1/xln(1+3x) = lim_(xrarroo)(ln(1+3x))/(x)#

This limit is of the indeterminate form #oo/oo# so use L'Hopital's.

#lim_(xrarroo)(ln(1+3x))/x = lim_(xrarroo)(d/(dx)(ln(1+3x)))/(d/(dx)(x)) = lim_(xrarroo) (3/(1+3x)) = 0#

Hence the limit of the exponent is 0 so overall limit is #e^0=1#

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