Limits of Powers

All of the different forms of powers of limits are handled in the same way. Since these functions don't have any obvious fractions in them, it doesn't look like l'Hôpital's Rule will apply at all to them. However, with the use of the exponential function, we can put a function into fractional form. Recall that

a^b = exp(b*ln(a))

and use that to find the limit. And, since the exponential function is continuous, you can use the Composition Limit Law to bring the limit inside the exponential function. In general, handle exponential limits in this way:

the limit of f(x)^(g(x)) = the limit of exp(g(x)*ln(f(x))) =

the limit of exp((ln(f(x)))/(1/g(x))) = exp(the limit of (ln(f(x)))/(1/g(x)).

Consider the following limit:

the limit as x goes to 0 from the right of x^x.

Using the regular limit laws, we cannot find this limit. Use the exponential function to change the form of your limit.

the limit as x goes to 0 from the right of x^x = the limit as x goes to 0 from the right of exp(x*ln(x))

Now we can use the Composition Limit Law to bring the limit "inside" the exponential function:

the limit as x goes to 0 from the right of exp(x*ln(x)) = exp(the limit as x goes to 0 from the right of x*ln(x))

but only if we can find a limit for the inside function. Now, we can rewrite the limit as follows:

the limit as x goes to 0 from the right of x*ln(x) = the limit as x goes to 0 from the right of (ln(x))/(1/x)

and since both ln(x) and 1/x have infinite limit, we can use l'Hôpital's Rule on the limit.

right-handed limit as x goes to 0 of (ln(x))/(1/x) *= right-handed limit as x goes to 0 of (1/x)/(-1/x^2) = right-handed limit as x goes to 0 of (-x^2)/x = right-handed limit as x goes to 0 of -x = 0

So,

the limit as x goes to 0 from the right of x^x = e^0 = 1.

Here is another example of how this method can work.

the limit as x goes to 0 of (cos(x))^(1/x^2) = the limit as x goes to 0 of exp((ln(cos(x)))/(x^2)) = exp(the limit as x goes to 0 of (ln(cos(x)))/(x^2))

Now, we can use l'Hôpital's Rule on the fraction, since both the numerator and denominator have limit zero,

the limit as x goes to 0 of (ln(cos(x)))/(x^2) *= the limit as x goes to 0 of (1/(cos(x))*(-sin(x)))/(2x) = the limit as x goes to 0 of -(tan(x))/(2x)

and then use it again to find the limit.

the limit as x goes to 0 of -(tan(x))/(2x) *= the limit as x goes to 0 of (-(sec^2(x))/2) = -(sec^2(0))/2 = -(1^2)/2 = -1/2

Now we can state that

the limit as x goes to 0 of (cos(x))^(1/(x^2)) = e^(-1/2) = 1/(the square root of e).

Here is a third example.

the limit as x goes to 1 from the right of (1/(x-1))^(ln(x)) = the limit as x goes to 1 from the right of exp((ln(x)*(-ln(x-1))) =

exp(the limit as x goes to 1 from the right of -(ln(x-1))/(1/ln(x)))

Now apply l'Hôpital's Rule to the fraction, as both the top and the bottom of the fraction have infinite limits.

the right-handed limit as x goes to 1 of -(ln(x-1))/(1/ln(x)) *= the right-handed limit as x goes to 1 of -(1/(x-1))/((-1/(ln^2(x)))*1/x) = the right-handed limit as x goes to 1 of (x*ln^2(x))/(x-1)

We can use l'Hôpital's Rule again here, this time with both limits being zero.

the limit as x goes to 1 from the right of (x*ln^2(x))/(x-1) *= the limit as x goes to 1 from the right of (1*ln^2(x)+2*x*ln(x)*(1/x))/1 =

the limit as x goes to 1 from the right of ln^2(x)+2ln(x) = 0^2+2*0 = 0

So, we can put the limit back in to find the final answer.

the limit as x goes to 1 from the right of (1/(x-1))^(ln(x)) = e^0 = 1

Copyright © 1996 Department of Mathematics, Oregon State University

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