Find $ \lim_{(x,y)\to(0,0)} \frac{\sin( |x| + |y|) + |y|(e^x - 1)} {|x| + |y|} $

Question

$$ \lim_{(x,y)\to(0,0)} \frac{\sin( |x| + |y|) + |y|(e^x - 1)} {|x| + |y|} $$ I tried in this way

$ \lim_{(x,y)\to(0,0)} \frac{\sin( |x| + |y|)}{|x| + |y|} + \frac{|y|(e^x - 1)} {|x| + |y|} $

the first term, when $(x,y)\to(0,0)$, is $1$. When $x\to 0 $ we have that $(e^x - 1) \to x$.

Now the limit to solve is: $\lim_{(x,y)\to(0,0)} 1 + \frac{|y|x} {|x| + |y|} = \lim_{(x,y)\to(0,0)} f(x,y) $

$ f(x,y) ≤ | \frac{|y|x} {|x| + |y|} |$ = $ \frac{|y||x|} {|x| + |y|} $ = $ \rho \frac{|\sin(\theta)||\cos(\theta)|}{|\sin(\theta)| + |\cos(\theta)|} $ ≤ $ \rho \frac{1}{2m} $ where $\frac{1}{2}$ is the maximum of the function in the numerator and m is the minimum of the function in the denominator and it is a positive number

$ \rho \frac{1}{2m} \to 0 $ when $\rho \to 0^+$ So the initial limit is 1

Is it ok?

Answer 1

$e^{x}-1 \to x$ is not a precise statement and it is actually much easier to handle the second term. Just note that $\frac {|y|} {|x|+|y|} \leq 1$ so the second term is bounded in absolute value by $|e^{x}-1|$ which tends to $0$.

Answer 2

For the second one we have

$$ \frac{|y|(e^x - 1)} {|x| + |y|}= \frac{e^x - 1} {x} \cdot\frac{|y|x} {|x| + |y|} \to 0$$

as you have shown by polar coordinates or as an alternative by AM-GM since $|x| + |y| \ge 2\sqrt{|xy|}$

$$\frac{|xy|} {|x| + |y|}\le \frac{|xy|} {2\sqrt{|xy|}}=\frac12 \sqrt{|xy|} \to 0$$