2
$\begingroup$

I am given with the function: $ f(x,y) = \frac{y\ln(1+x^2 + ay^2) } {x^2 + 2y^2} $ when $ (x,y)\neq (0,0)$, and $f(0,0)=0$ .

There is another given data; $ f_y (0,0) = 2 $ . What is the value of $a$ ?

I've tried computing the limit $ \frac{f(0,h)- f(0,0)}{h} $ , but it seems like it's always zero, contradicting the fact that $f_y(0,0)=2 $ !

Can someone help me understand my mistake?

Thanks !

  • 1
    A number followed by an exclamation mark reminds me of factorial5!2012-07-21

2 Answers 2

2

$ \begin{align} f_y(0, 0) &= \lim_{h \to 0} \frac {f(0, h) - f(0, 0)} h\\ &= \lim_{h \to 0} \frac {f(0, h)} h\\ &=\lim_{h \to 0} \frac a 2 \frac {\ln(1 + ah^2)} {ah^2}\\ &= \lim_{h \to 0} \frac a 2 \frac {\ln(1 + h)} {h}\\ &= \frac a 2 \left. \frac {d} {dx} \right |_{x=1} \ln x\\ &= \frac a 2 \end{align} $ Since $f_y(0, 0) = 2$, $a$ must be $4$.

  • 0
    @joshua: $\displaystyle \frac {f(0, h)} h = \frac 1 h \frac {h \ln(1 + ah^2)} {2h^2} = \frac {\ln(1 + ah^2)} {2h^2} = \frac a 2 \frac {\ln(1 + ah^2)} {ah^2}$2012-07-21
0

What about a direct L'Hospital (=L'H)?:

$\lim_{h\to 0}\frac{1}{h}f(0,h)=\lim_{h\to 0}\frac{1}{\rlap{/}h}\frac{\rlap{/}h\log(1+ah^2)}{2ah^2}\stackrel{\text{L'H}}=\lim_{h\to 0}\frac{\rlap{/}2a\rlap{/}h}{1+ah^2}\frac{1}{\rlap{/_2}4\,\,\,\rlap{/}h}=\frac{a}{2}$