# What is analysis and why do analysis — part 2 of 2

We had discussed this on Nov 17 2015 blog. We finish the article with more examples from the work of Prof. Terence Tao. (If you like it, please send a thanks to him :-))

Example 1. (Interchanging limits and integrals).

For any real number y, we have $\int_{-\infty}^{\infty}\frac {dx}{1+(x-y)^{2}}=\arctan (x-y)\mid_{x=-\infty}^{\infty}$ which equals $(\pi/2)-(-\pi/2)=\pi$.

Taking limits as $y \rightarrow \infty$, we should obtain $\int_{-\infty}^{\infty}\lim_{y \rightarrow \infty}\frac {dx}{1+(x-y)^{2}}=\lim_{y \rightarrow \infty} \int_{-\infty}^{\infty}\frac {dx}{1+(x-y)^{2}}=\pi$

But, for every x, have $\lim_{y \rightarrow \infty} \frac {1}{1+(x-y)^{2}}=0$. So, we seem to have concluded that $0=\pi$. What was the problem with the above argument? Should one abandon the (very useful) technique of interchanging limits and integrals?

Example 2. Interchanging limits and derivatives.

Observe that if $\in > 0$, then $\frac {d}{dx}\frac {x^{3}}{\in^{2}+x^{2}}=\frac {3x^{2}(\in^{2}+x^{2})+x^{2}-2x^{4}}{(\in^{2}+x^{2})^{2}}$,

and in particular that $\frac {d}{dx}\frac {x^{3}}{\in^{2}+x^{2}}\mid_{x=0}=0$.

Taking limits as $\in \rightarrow 0$, one might then expect that $\frac {d}{dx}\frac {x^{3}}{0+x^{2}}\mid_{x=0}=0$.

But, the right hand side is $\frac {dx}{dx}=1$. Does this mean that it is always illegitimate to interchange limits and derivatives?

Example 3. Interchanging derivatives. $Let^{1}$ $f(x,y)$ be the function $f(x,y)=\frac {xy^{3}}{x^{2}+y^{2}}$. A common manoeuvre in analysis is to interchange two partial derivatives, thus one expects $\frac {\partial^{2}f(0,0)}{\partial x \partial y}=\frac {\partial^{2}f(0,0)}{\partial y \partial x}$.

But, from the quotient rule, we have $\frac {\partial f(x,y)}{\partial y}=\frac {3xy^{2}}{x^{2}+y^{2}}=\frac {2xy^{4}}{(x^{2}+y^{2})^{2}}$

and in particular, $\frac {\partial f(x,0)}{\partial y}=\frac {0}{x^{2}}-\frac{0}{x^{4}}=0$.

Thus, $\frac {\partial^{2}f(0,0)}{\partial x \partial y}=0$.

On the other hand, from the quotient rule again, we have $\frac {\partial f(x,y)}{\partial x}=\frac {y^{3}}{x^{2}+y^{2}} - \frac {2x^{2}y^{3}}{(x^{2}+y^{2})^{2}}$ and hence, $\frac {\partial f(0,y)}{\partial x}=\frac {y^{3}}{y^{2}}-\frac {0}{y^{4}}=y$.

Thus, $\frac {\partial^{2}f(0,0)}{\partial y \partial x}=1$.

Since $1 \neq 0$, we thus seem to have shown that interchange of two derivatives is untrustworthy. But, are there any other circumstances in which the interchange of derivatives is legitimate?

Example 4. $L^{'} H\hat {o}pital's Rule$

We are familiar with the beautifully simple $L^{'}H \hat{0}pital's$ rule $\lim_{ x \rightarrow x_{0}} \frac {f(x)}{g(x)}=\lim_{x \rightarrow x_{0}}\frac {f^{'}(x)}{g^{'}(x)}$

but one can still get led to incorrect conclusions if one applies it incorrectly. For instance, applying it to $f(x)=x$, $g(x)=1+x$ and $x_{0}=0$ we would obtain $\lim_{x \rightarrow 0}\frac {x}{1+x}=\lim_{x \rightarrow 0} \frac {1}{1}=1$.

But this is an incorrect answer since $\lim_{x \rightarrow 0}\frac {x}{1+x}=\frac {0}{1+0}=0$.

Of course, all that is going on here is that $L^{'}H \hat{o}pital's rule$ is only applicable when both $f(x), g(x)$ go to zero as $x \rightarrow x_{0}$, a condition which was violated in the previous example. But, even when $f(x)$ and $g(x)$ do go to zero as $x \rightarrow x_{0}$, there is still a possibility for an incorrect conclusion. For instance, consider the limit $\lim_{x \rightarrow 0} \frac {x^{2} \sin (x^{-4})}{x}$.

Both numerator and denominator go to zero as $x \rightarrow 0$, so it seems pretty safe to apply the rule, to obtain $\lim_{x \rightarrow 0} \frac {x^{2}\sin (x^{-4})}{x}=\lim_{x \rightarrow 0} \frac {2x \sin (x^{-4})-4x^{-3}\cos (x^{-4})}{1}$ which equals $\lim_{x \rightarrow 0}2x \sin (x^{-4})-\lim_{x \rightarrow 0}4x^{-3}\cos (x^{-4})$.

The first limit converges to zero by the Sandwich theorem (since the function $2xsin(x^{-4})$ is bounded above by $2|x|$ and below by $-2|x|$, both of which go to zero at 0). But the second limit is divergent (because $x^{-3}$ goes to infinity as $x \rightarrow 0$, and $\cos (x^{-4})$ does not go to zero.) So the limit $\lim_{x \rightarrow 0} \frac {2x \sin(x^{-4})-4x^{-2}\cos (x^{-4})}{1}$ diverges. One might then conclude using $L^{'}H\hat{o}pital's Rule$ that $\lim_{x \rightarrow 0}\frac {x^{2}\sin (x^{-4})}{x}$ also diverges; however, we can clearly rewrite this limit as $\lim_{x \rightarrow 0}x\sin(x^{-4})$, which goes to zero when $x \rightarrow 0$ by the Sandwich Theorem again. This does not show that $L^{"}H\hat opital's Rule$ is untrustworthy. Indeed, it is quite rigorous, but it still requires some care when applied.

That is all, once again, if you like this, please send a thanks note to Prof. Terence Tao.

More later,

Nalin Pithwa

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