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What is the limit of #x^(1/x)# as #x->oo#?

Answer 1

Scarlett Allison

#Lt_(x->oo)x^(1/x)=1#

Let #x^(1/x)=e^a#, then taking natural log on both sides, we get

#1/xlnx=a# and #x^(1/x)=e^(lnx/x)#

Hence, #Lt_(x->oo)x^(1/x)=Lt_(x->oo)e^(lnx/x)#

and as #e^u# is continuous #Lt_(x->oo)x^(1/x)=e^(Lt_(x->oo)(lnx/x))#

Now in #Lt_(x->oo)e^(lnx/x)# as #x->oo#, both numerator and denominators tend to infinity,

therefore we can apply L'Hospital's Rule and

#Lt_(x->oo)(lnx/x)=Lt_(x->oo)((1/x)/1)=Lt_(x->oo)1/x=0#

Hence #Lt_(x->oo)x^(1/x)=e^0=1#

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Answer 2

Emilia Aguilar

The limit of x^(1/x) as x approaches infinity is 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

What is the limit of #x^(1/x)# as #x-&gt;oo#?

What is the limit of #x^(1/x)# as #x->oo#?