Question

1. show that the limit does not exist

\\\lim_{(x,y) \to (0,0)} \frac{x + y}{2x - y}\\

Explanation:

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<reasoning>

Evaluate the limit along the path \(y = 0\)

\[
\lim_{(x,0) \to (0,0)} \frac{x + 0}{2x - 0} = \lim_{x \to 0} \frac{x}{2x} = \frac{1}{2}
\]

Evaluate the limit along the path \(x = 0\)

\[
\lim_{(0,y) \to (0,0)} \frac{0 + y}{2(0) - y} = \lim_{y \to 0} \frac{y}{-y} = -1
\]

Compare the limits along different paths

\[
\frac{1}{2}
eq -1
\]
</reasoning>

<answer>
The limit does not exist because evaluating the limit along two different paths yields different values:

• Along the path \(y = 0\), the limit is \(\frac{1}{2}\).
• Along the path \(x = 0\), the limit is \(-1\).

Since the limit depends on the path of approach, \(\lim_{(x,y)\to(0,0)} \frac{x+y}{2x-y}\) does not exist.
</answer>

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Answer:

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<reasoning>

Evaluate the limit along the path \(y = 0\)

\[
\lim_{(x,0) \to (0,0)} \frac{x + 0}{2x - 0} = \lim_{x \to 0} \frac{x}{2x} = \frac{1}{2}
\]

Evaluate the limit along the path \(x = 0\)

\[
\lim_{(0,y) \to (0,0)} \frac{0 + y}{2(0) - y} = \lim_{y \to 0} \frac{y}{-y} = -1
\]

Compare the limits along different paths

\[
\frac{1}{2}
eq -1
\]
</reasoning>

<answer>
The limit does not exist because evaluating the limit along two different paths yields different values:

• Along the path \(y = 0\), the limit is \(\frac{1}{2}\).
• Along the path \(x = 0\), the limit is \(-1\).

Since the limit depends on the path of approach, \(\lim_{(x,y)\to(0,0)} \frac{x+y}{2x-y}\) does not exist.
</answer>

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