0
$\begingroup$

My teacher gave us this problem where we need to find:

$$\lim_{(x,y) \to (0,0)} \frac{1-cos(x+y)}{x+y}$$

My first gut instinct would have been to try sandwich theorem (after attempting a few paths and getting all $0$s). However, he gave us a solution using a Taylor approximation:

$$\lim_{(x,y) \to (0,0)} \frac{\frac1 2 (x^2+2xy+y^2)+R_2(x,y)}{x+y} = \lim_{(x,y) \to (0,0)} \frac{\frac1 2 (x+y)^2}{x+y}+ \frac{R2(x,y)}{x+y}=0$$

It seems he used a second degree Taylor polynomial with its remainder term (which tends to $0$ when the function and the Taylor polynomial are close to the same point).

However, three questions arise:

1) Can I always use this Taylor approach to avoid taking a limit by using paths and sandwich theorem? If not, when exactly can or cannot I use Taylor for limits?

2) Why a second degree polynomial? How should I know which grade to use in this technique?

3) I don't get where the $2xy$ came from in the polynomial (inside the parentheses) and why some terms are positive instead of negative. Since $\cos(x+y)$ equals $1$ when evaluated at $(0;0)$, the first derivatives evaluated at $(0;0)$ are equal to $0$ and the second derivatives are all equal to $-1$ then, to me, the polynomial should look like: $$ 1+\frac 1 2 (-x^2-xy-y^2)+R_2(x,y) $$

$\endgroup$
  • $\begingroup$ the searched limit is zero $\endgroup$ – Dr. Sonnhard Graubner Dec 10 '16 at 13:19
  • $\begingroup$ Thanks, that's the same results our teacher gave us. However, I fail to understand the steps he took in his solution. $\endgroup$ – Floella Dec 10 '16 at 13:32
  • $\begingroup$ The answer to the first question is yes you can replace it by approximating series, but in multivariate calculus you always have to choose paths and check the limits. (look the example I posted with a generic path). When the limit goes to 0, polynomials of higher order become unnecessary, compute only up to the necessary degree. The $2xy$ term comes from $(x+y)^2$ in the Taylor series of $\cos$; or if you don't see it that way, check out that $\cos(x+y)$ is a two variable function and has a Taylor expansion: $$f(x,y)\approx1+(\partial_xf)x+(\partial_y f)y + (\partial_x\partial_yf)xy $$ $\endgroup$ – HBR Dec 10 '16 at 13:40
0
$\begingroup$

This limit can be computed as follows $$\lim_{(\vec{r}\cdot\vec{u})\to 0}\frac{1-\cos{(\vec{r}\cdot\vec{u})}}{(\vec{r}\cdot\vec{u})}\tag1$$ With $\vec{u}$ an arbitrary unitary vector and $\vec{r}$ the position vector both of them 2 dimensional. Then you can expand the $\cos{(\vec{r}\cdot\vec{u})}$ in Taylor series $$\cos{(\vec{r}\cdot\vec{u})}\approx 1-\frac{1}{2}(\vec{r}\cdot\vec{u})^2+\mathcal{o}((\vec{r}\cdot\vec{u})^2)$$ Substituting back in $(1)$ $$\lim_{(\vec{r}\cdot\vec{u})\to 0}\frac{1-\cos{(\vec{r}\cdot\vec{u})}}{(\vec{r}\cdot\vec{u})}=\lim_{(\vec{r}\cdot\vec{u})\to 0}\left[(\vec{r}\cdot\vec{u})+\mathcal{o}((\vec{r}\cdot\vec{u}))\right]=\lim_{r\to 0}r\cos{\theta}=0$$ whatever $\theta$ we take

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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