Limits
Task number: 3193
Find the following limits, or justify why they don't exist.
Variant 1
\( \lim\limits_{(x,y) \to (0{,}2)} \frac{\sin xy}{x}. \)
Resolution
Even though at first look it may seem that the limit is 2, the limit does not exist. The reason is that the function is not defined on any annular neighborhood of the point \((0, 2)\). (Specifically, the function is not defined if \( x = 0 \) and \( y \) near \( 2 \).)
Variant 2
\( \lim\limits_{(x,y) \to (0{,}0)} \frac{\sin (x^2 + y^2)}{x^2 +y^2}. \)
Hint
First of all, we note that the function is defined at all points except at the point \((0, 0)\), because \(x^2 + y^2 = 0\) only for \((x, y) = (0, 0)\). Thus the function is defined at some annular neighbourhood of \((0, 0)\), which is the necessary condition for a limit to exist.
We will use the theorem of limits for composite functions (for several variables).
Define the function \(f(x, y) = x^2 + y^2\) and \(g(t) = \frac{\sin t}{t}\). We calculate the limit of
\[ \lim\limits_{(x,y) \to (0{,}0)} g(f(x, y)). \] We know that \[ \lim\limits_{(x,y) \to (0{,}0)} f(x, y) = 0, \] since \(f\) is continuous, and also \[ \lim\limits_{t \to 0} g(t) = 1 \] (a well known limit in a single variable).
The multivariable limit of a composite function uses analogous assumptions as in one variable. We either need the outer function to be continuous (which it is not), or the inner function does not assume the limit values, ie \( 0 \), on some annular neighborhood of the point at which we calculate the limit, ie, the \( (0, 0) \). But the second condition is fulfilled, as we realized at the beginning. Thus, according to the theorem of the composite function limit:
\[ \lim\limits_{(x,y) \to (0{,}0)} \frac{\sin (x^2 + y^2)}{x^2 +y^2} = 1. \]
Variant 3
\( \lim\limits_{(x,y) \to (0{,}0)} \frac{x^2 - y^2}{x^2 +y^2}. \)
Hint
The limit does not exist. Let \(f(x, y) = \frac{x^2 - y^2}{x^2 + y^2}\). As we approach \((0, 0)\) with the help of the sequence \((1/n, 0)\) we get \[ \lim\limits_{n \to \infty} f(1/n, 0) = 1. \] However, as we approach the point \((0, 0)\) with the help of the sequence \((1/n, 1/n)\) we get \[ \lim\limits_{n \to \infty} f(1/n, 1/n) = 0. \] Since the limit must be single valued, it cannot exist.
Variant 4
\( \lim\limits_{(x,y) \to (0{,}0)} \frac{\ln (1+xy)}{|x| + |y|}. \)
Resolution
We know that \[\lim_{x\to 0}\frac{\ln(1+t)}{t}=1.\] This leads us to decompose the expression into the product \[ \frac{\ln(1+xy)}{xy}\cdot \frac{xy}{|x| + |y|} \] and try to calculate \[\lim_{(x,y)\to (0{,}0)}\frac{\ln(1+xy)}{xy}\] and \[ \lim_{(x,y)\to (0{,}0)}\frac{xy}{|x|+|y|}. \]
It may seem, according to the first equation, that this limit could be equal to \(1\). However, as we saw in a previous problem (Variant 1), the limit does not exist because the function is not defined in an annular neighborhood of the point \((0, 0)\). We will not let this put us off and we will pretend that the first limit is equal to \(1\). We will show later what changes have to be made in order to get the right answer.
Let us now calculate the second limit. The limit exists and can be calculated (using a small trick)
We use the estimate \[ 0\le \frac{|xy|}{|x|+|y|}= |x|\frac{|y|}{|x|+|y|}\le |x|. \] Using an analogy of the "policeman lemma" (or the "squeeze theorem" ) we get \[ \lim_{(x,y)\to(0{,}0)} \frac{|xy|}{|x|+|y|} = 0. \]
At this point we would guess that the limit to the given problem will turn out to be \(1 \, \cdot \, 0 = 0\), however we still have to deal with the problem of non existence of \[\lim_{(x,y)\to (0{,}0)}\frac{\ln(1+xy)}{xy}.\]
In the neighborhood of \((0, 0)\) we consider two cases:
If \(xy = 0\), then \(\ln(1 +xy) = 0\) and the function in the given problem is identically \(0\).
If \(xy \neq 0\), then the function \(\ln(1 + xy)/xy\) is defined. Since \[\lim_{t \to 0} \frac{\ln(1 + t)}t = 1,\]
we get that there exists a \(\delta > 0\) such that \[|\ln(1+t)/t| \leq 2\] for all \(t \neq 0\) in a \(\delta\)-neighborhood of \(0\) (we get the \(\delta\) by selecting \(\varepsilon = 1\) in the definition of the limit). Combining the two cases, we get \[ \frac{|\ln(1 +xy)|}{|x|+|y|} \leq \frac{2|xy|}{|x|+ |y|}, \] as long as \(|xy| < \delta\). We can now safely make the claim that the limit of the given problem is \(0\).
