Thread: Lim x^{1/x}, x-> infty View Single Post
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 Quote by ice109 i don't understand your proof, but what i mean is that as the expression stands $$\frac{ln(x)}{x}$$ goes the the undeterminate form $$\frac{\infty}{\infty}$$ and that l'hospitale has to be applied to see that the limit is zero. the initial point was that the rewriting of the expression using logs didn't save him any steps because still had to apply l'hospitale to figure out the limit. i then realized that applying l'hospitale to original expression was pointless.

$$\frac{ln(x)}{x}$$ does in fact go to $$\frac{\infty}{\infty}$$ when x tend to inf.
this is an indeterminate form. u can use l'hopital's rule to solve this easily.

easier way of doing this can be explained usingthe following example
lim x -->inf (x^5/x^4). lim x--> inf (x^5) is inf and lim x-->inf (x^4) is inf. it is inf because x^5 grows faster than x^4 when x tends to infinity.

so the higher the power of x, the faster it grows. one thing we have to understand is that e^x grows faster than ANY function of x and ln(x) grows slower than ANY function of x when x tends to infinity. so $$\frac{ln(x)}{x}$$ when x tends to inf is just zero.

so both answers of indeterminate and zero is correct. inf/inf is just a way of expressing two functions are going towards infinity. it is an indeterminate value because we do not know how fast each functions are going towards infinity. and inf/inf is NOT always 1. in the example given by HallsofIvy. "limx->inf (x/x) is not inf/inf but 1." fact is it CAN BE expressed as inf/inf and l'hopital's rule can be used, which would give the answer 1. and this example gives an answer of 1 only because we know the two functions grows at the same rate (simply because they are the same function). so if the example was 2x/x. this would also be inf/inf when x-->inf. but the answer would not be 1 because the two functions are NOT growing at the same rate. and also simple questions like lim x-->inf (2x/x) would not require l'hopital's rule because we can instantly see that 2x grows twice as fast as x, or can easily simplify this with algebra. BUT using l'hopital's rule for this is perfectly fine.

i hope this clears up the argument between the answers of zero and inf/inf. they are both correct.