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can anyone give an example of functions f and g such that
lim x-> inf of f/g = 1 but lim x->inf of e^f/e^g is not 1?

That's what 29 is saying, right? Thanks.

f = sin(x)
g = x

f/g tends to 1 (check with lhospitals rule)

Edit: Sorry I was thinking as x tended toward 0 Ignore everything .

e^f oscillates between [1/e,e]
x^sin(x) oscillates as well but with increasing amplitude end up going from [0,Infinity]

thus e^f / g^f does not even converge as x tends to infinity

thanks for all your help, moo.

I'm still not quite understanding this. I had a typo in my original question, which I've fixed, maybe that caused some confusion.

the lim as x-> inf of sin(x)/x goes to 0, not 1, right? (sin is bounded, x gets arbitrarily large).

lim_(x----->0)(sinx)/x=1 . This one is very famous, you can prove it using L'Hopistal rule

thanks everyone. however, this problem talks about the limit as x goes to infinity, not zero.

Is there any way one can just "tell" that the answer is C, or do you find a specific counterexample where C is not implied from the problem?

Okie

let f(x)=x^2+x
g(x)=x^2

then f/g------>1 when x------>infinity
e^f/e^g=e^(f-g)=e^x------->infinity when x------>infinity

nice. Thanks!

A follow-up to this question: why is (D) always correct?

lim(x->inf) (f/g) = 1 always implies lim(x->inf) ((f+g)/2g) = 1

Can someone explain the above please? Thanks...

lim(x->inf) (f/g) = 1
then
lim(x->inf) ((f+g)/2g) = lim(x->inf) (f/2g+g/2g)=1/2+1/2=1

It's just algebra! Thanks Nameless! This test really requires a cool mind to see the problem from different angles - especially under exam stress.