Find $$\lim_{(x,y)\to(0,0)} \frac{e^{\sin(x^2+y^2)}-1}{x^2+y^2}$$
I have tried taking limits when $y=x$ and when $y=-x$ but failing to get close to an answer.
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2Can you see that it's going to be the same as $\lim_{p\to0}\frac{e^{\sin p}-1}{p}$? – πr8 Apr 30 '16 at 08:58
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2Welcome to the site ! By the way, where are you located ? Your profile does not say much. Cheers. – Claude Leibovici Apr 30 '16 at 08:59
1 Answers
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- Method 1. One may write, as $x^2+y^2 \to 0$, by the use of Taylor series expansions, $$ \begin{align} \sin(x^2+y^2)&=x^2+y^2+O((x^2+y^2)^3) \\\\e^{\sin(x^2+y^2)}&=1+x^2+y^2+O((x^2+y^2)^2) \end{align} $$ giving, as $x^2+y^2 \to 0$, $$ \frac{e^{\sin(x^2+y^2)}-1}{x^2+y^2}=1+O(x^2+y^2). $$
- Method 2. One may write, as $x^2+y^2 \to 0$, $$ \frac{e^{\sin(x^2+y^2)}-1}{x^2+y^2}=\frac{\color{red}{e^{\sin(x^2+y^2)}-1}}{\color{red}{\sin(x^2+y^2)}}\times \frac{\color{blue}{\sin(x^2+y^2)}}{\color{blue}{x^2+y^2}}. $$
Can you take it from here?
Olivier Oloa
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Thank you so much for the help, managed to get an answer using your 2nd method. :) – Andre Croucher Apr 30 '16 at 09:10
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