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

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.

• Hint: $f(a,b)=\max \{a,b\}$. – Kavi Rama Murthy May 20 '20 at 12:29
• I didn't get you @Kavi Rama Murthy – Stranger Forever May 20 '20 at 12:34

## 2 Answers

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 $$a is similar. I will let you check this when $$a=b$$].

• 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 $(c)$? – Stranger Forever May 20 '20 at 13:19
• please see my comment.. – Stranger Forever May 20 '20 at 13:47
• $f$ 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 $x=b$. – Kavi Rama Murthy May 20 '20 at 23:20
• 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.. – Stranger Forever May 21 '20 at 3:21
• 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 – Kavi Rama Murthy May 21 '20 at 5:02

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 $$f(a,b) =\max\{a,b\}$$.