limit calculation with x __>`o o `

limit x ^(1/x^.5)

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You need to evaluate the limit of the function` lim_(x->oo) x^(1/sqrt x)` , hence, you need to substitute oo for x in equation of function, such that:

`lim_(x->oo) x^(1/sqrt x) = oo^(1/oo) = oo^o`

You may use the following approach to evaluate this limit, such that:

`lim_(x->oo) x^(1/sqrt x) = lim_(x->oo) e^(ln(x^(1/sqrt x)))`

Using the limits properties yields:

`lim_(x->oo) e^(ln(x^(1/sqrt x)) = e^(lim_(x->oo)(ln x^(1/sqrt x))))`

Using logarithmic power identity yields:

`lim_(x->oo)ln(x^(1/sqrt x)) = lim_(x->oo) (1/sqrt x)*ln x`

`lim_(x->oo) (1/sqrt x)*ln x = (ln oo)/oo = oo/oo`

The indetermination case `oo/oo` requests for you to use l'Hospital's theorem, such that:

`lim_(x->oo) (ln x)/sqrt x = lim_(x->oo) ((ln x)')/((sqrt x)')`

`lim_(x->oo) ((ln x)')/((sqrt x)') =` `lim_(x->oo) ((2sqrt x)/x)`` `

Reducing duplicate factors yields:

`lim_(x->oo) ((2sqrt x)/x) = lim_(x->oo) 2/(sqrt x) = 2/oo = 0`

Replacing 0 for `lim_(x->oo)ln(x^(1/sqrt x))` yields:

`e^(lim_(x->oo)ln(x^(1/sqrt x))= e^0 = 1`

**Hence, evaluating the given limit, using the indicated approach yields `lim_(x->oo) x^(1/sqrt x) = 1` .**

`lim_(x->oo)``x^(1/sqrt(x))`

Taking the limit results in the indeterminite form of `oo^0`

With this you want to transform it into the form of `oo/oo`

So, take the natural log of it we will get `e^(lim_(x->oo)lnx^(1/sqrt(x)))`

Simplifying this, we get `e^(lim_(x->oo))ln(x)/sqrt(x)` which is in the form of `oo/oo`

Now we can use L'Hospital's Rule to get:

`e^(lim_(x->oo)((lnx)')/((sqrtx)'))` = `e^(lim_(x->oo)2/sqrtx)`

So, if we just look at the limit we see that `lim_(x->oo)2/sqrtx=0`

Finally, we plug in 0 for the whole limit, making the equation

`e^0` which is equal to 1

Hence, `lim_(x->oo)x^(1/(sqrtx))=1`

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