Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 !

share|improve this question
1  
A number followed by an exclamation mark reminds me of factorial5! –  FrenzY DT. Jul 21 '12 at 8:42
add comment

2 Answers 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 $4$.

share|improve this answer
    
Dear @Albert: There's a fallacy in what you wrote. When substituting y=h, you don't get $ \frac{a}{2} $ outside the fraction, but $\frac{h}{2 }$ ! In addition, you'll get in the denominator $2h^2 $ and not $ah^2$ . –  joshua Jul 21 '12 at 8:49
    
@joshua: $$\displaystyle \frac {f(0, h)} h = \frac 1 h \frac {h \ln(1 + ah^2)} {2h^2} = \frac {\ln(1 + ah^2)} {2h^2} = \frac a 2 \frac {\ln(1 + ah^2)} {ah^2}$$ –  AlbertH Jul 21 '12 at 8:58
    
OH right... Thanks a lot !!! –  joshua Jul 21 '12 at 9:13
    
You are welcome! –  AlbertH Jul 21 '12 at 15:07
add comment

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

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.