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QUESTION: Consider $f :\Bbb B\times\Bbb R\to\Bbb R$ defined as follows: $$f(a,b) := \lim_{n\to\infty} \frac{1} n\ln[e^{na}+ e^{nb}]$$

Then state which of the following is true or false-

$(a)$ $f$ is not onto i.e. the range of $f$ is not all of $\Bbb R$.

$(b)\ \forall a$ the function $x\mapsto f(a,x)$ is continuous everywhere.

$(c)\ \forall b$ the function $x\mapsto f(x,b)$ is differentiable everywhere.

$(d)$ We have $f(0,x) = x\ \forall x\geqslant 0$.


MY APPROACH: I tried to calculate the limit. Since it is in $\frac{\infty}\infty$ form, we can use L'Hospitals rule here. I applied the same. But the problem is after calculation, it comes out as- $$\lim\limits_{n\to\infty}\frac{ae^{na}+be^{nb}}{e^{na}+e^{nb}}$$ Now, since the exponential function is infinitely differentiable I could not come to a solution. I cannot even cancel any term from the numerator and the denominator. Everytime I try to differentiate another coefficient multiplies infront of every term. Then I tried to divide both the numerator and the denominator by $e^{na}$ but that again failed to help.

Coming to the options, option $d$ is easy. If I calculate the limit of $f(0,x)$ then I arrive at $$\lim_{nto\infty}\frac{\ln({1+e^{nx}})}{n}$$ which after applying the L'Hospitals rule we get- $$\lim_{n\to\infty}\frac{ne^{nx}}{1+e^{nx}}$$ Now, this is trivial and after dividing the numerator and the denominator by $e^{nx}$ we easily see that the limit is indeed equal to $x$. I hope I am correct.

But what about the rest? How do I solve them out?

Any help will be much appreciated. Thank you so much.

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  • $\begingroup$ Hint: $f(a,b)=\max \{a,b\}$. $\endgroup$ – Kavi Rama Murthy May 20 '20 at 12:29
  • $\begingroup$ I didn't get you @Kavi Rama Murthy $\endgroup$ – Stranger Forever May 20 '20 at 12:34
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Let $a >b$. Then $$\log_e[e^{na}+e^{mb}]=\log_e [(e^{na}) (1+e^{-n(a-b)})]$$ $$=\log_e e^{na}+\log_e[1+e^{-n(a-b)}]$$ $$=na+\log_e[1+e^{-n(a-b)}]$$. Use the fact that $\log (1+x) \sim x$ for $|x| \to 0$ to show that $f(a,b)=\max \{a,b\}$. [ The case $a<b$ is similar. I will let you check this when $a=b$].

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  • $\begingroup$ So we may conclude that $f$ is a constant function isn't it? And the range of $f$ can be any real number depending on the value of $a$. That makes option $(a)$ false.. now, since $f$ is a constant function, it is obviously continuous, so we have option $(b)$ correct. But what about option $(c)$? $\endgroup$ – Stranger Forever May 20 '20 at 13:19
  • $\begingroup$ please see my comment.. $\endgroup$ – Stranger Forever May 20 '20 at 13:47
  • $\begingroup$ $f$ is not a constant function. $a$ and $b$ are variables and $f(a,b)=\max \{a.b\}$. a) and b) are true and c) and d) are false. For c) note that differentiability fails at $x=b$. $\endgroup$ – Kavi Rama Murthy May 20 '20 at 23:20
  • $\begingroup$ can you please show that to me. I did not get you how the differentiability fails. And why is $d$ false? I have shown that in my answer.. $\endgroup$ – Stranger Forever May 21 '20 at 3:21
  • $\begingroup$ Sorry, I mis-read d)$. d)$ is true. For c) note that the left-hand derivative at $b$ is $0$ and the right hand derivative is $1$. @StrangerForever $\endgroup$ – Kavi Rama Murthy May 21 '20 at 5:02
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Assuming $b\ge a$ without loss of generality, $$\lim_{n\to \infty} \frac 1n \ln(e^{na} + e^{nb} ) =\lim_{n\to\infty} \frac 1n \left(na + \ln\left(1+e^{n(b-a)}\right) \right) \\ =a+\lim_{n\to\infty}\frac{\ln\left(1+e^{n(b-a)} \right)}{n} \overset{\text{L.H.}}=a+b-a =b$$

Similarly, if $a\gt b$ then $f(a,b) =a$ and so $f(a,b) =\max\{a,b\}$.

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