Question

Why can't we apply L'Hospital's Rule to the

lim as x->a f(x) ^ g(x)

if f(x) = negative infinity and g(x) = 0

but it works if f(x) is infinity?

Answer 1

for one thing if the base is negative you are in a lot of trouble
the outputs will not be real numbers

Answer 2

I don't believe L'Hospital's can be done here ... I think it only works when there is a quotient of two functions and either both tend to +/- infinity or both tend to zero.

Answer 3

it is not a matter of l'hopital or no l'hopital you cannot have a negative base

Answer 4

Unless natural logarithms are taken.

Answer 5

not with real numbers

Answer 6

what, for example, would \(-10^{\frac{1}{2}}\) be?

Answer 7

the base of any exponential function has to be positive

Answer 8

That would be imaginary. But they can still be raised to integer powers.

Answer 9

logical reasoning, yes I get that part and taking the natural log

satellite, well its not defined in real numbers for a negative base to certain fractional powers but it is defined for many other numbers like (-2)^2 or -10^(1/3)

Answer 10

(-9)^(1/3) is -3 for example

Answer 11

i am not saying you cannot take a negative number to some powers of your choosing
i am saying that the base of any exponential function must be positive, or else it is not a real valued function
there is no such function as for example \(f(x)=(-3)^x\)

Answer 12

if you have a function with a variable in the exponent, that variable can take on infinitely many values,
not just integers or specific fractions.

Answer 13

in fact, how do we even interpret something like \(x^{\sin(x)}\)? the definition is given in terms of the logarithm, so \[x^{\sin(x)}=e^{\sin(x)\ln(x)}\] and is only defined for positive values of \(x\)

Answer 14

thats a good point