Mycroft Holmes is in the midst of solving a limit mystery caused by the ink washing off of an old priceless relic: a 20 year old MAO test. He knows that the limit is of the form $\lim_{x\rightarrow \infty }\frac{f(x)}{g(x)}$ and that f(x)and g(x)are distinct functions from the set {log(x^5), log(2^x),2x+sinx,x^3, and 2^(x+logx)}.Thinking in terms of his catchphrase,‘balance of probability', what is the probability that the limit works out to zero?
A: The first thing to note is that there are 5 x 4 = 20 possible assignments for distinct f(x) and g(x). Additionally, for every f(x), g(x) such that $\lim_{x\rightarrow \infty }\frac{f(x)}{g(x)}$ = 0, $\lim_{x\rightarrow \infty }\frac{f(x)}{g(x)}$ = ±∞ or DNE, meaning that exactly half of these limits that don’t evaluate to a nonzero finite value will equal zero. Thus, we are looking for the number of f(x), g(x) pairs such that this limit evaluates to a nonzero finite value. This is only true for f(x), g(x) pairs that are asymptotically equal save for a constant factor, and from the set of 5 equations, that is only the case for log(2x )= xlog2 and 2x+sinx, and there are 2 ways to match these two equations to f(x) and g(x). Thus, the number of distinct f(x), g(x) pairs for which the limit is equal to zero is 20 − 2 = 9 , meaning our final probability is 9/20
My questions are, why can g(x)/f(x) also be DNE? wouldn't f(x)/g(x) be DNE as well if that was the case? Also can someone explain to me why f(x) and g(x) have to be "asymptotically equal, save for a constant factor" to evaluate to a nonzero finite value? Also why is log(2x) and 2x+sinx asymptotically equal?
$f(x)/g(x)$ can go to zero with alternating signs, and then $g(x)/f(x)$ diverges because one subsequence goes to $\infty$ and another to $-\infty$.
The answer to your second question is that this is essentially the definition of “asymptotically equal”: that the ratio converges to $1$ (and thus, if you apply this “save for a constant factor”, the ratio converges to that constant factor).
The answer to your third question is that $\log2^x$ and $2x+\sin x$ are not in fact asymptotically equal. Their ratio does, however, converge to a non-zero constant:
$$ \lim_{x\to\infty}\frac{2x+\sin x}{\log2^x}=\lim_{x\to\infty}\left(\frac{2x}{x\log2}+\frac{\sin x}{x\log2}\right)=\lim_{x\to\infty}\frac{2x}{x\log2}+\lim_{x\to\infty}\frac{\sin x}{x\log2}=\frac2{\log2}+0=\frac2{\log2}\;. $$