Limit Solving Algorithm

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General Strategy

To evaluate limits of the form\[\lim_{(x,y)\to(a,b)} f(x,y)\]we follow a systematic decision process.

Step 1: Direct Substitution

Substitute the point directly.\[f(a,b)\]Cases

• Finite value → limit exists
• Indeterminate form → continue

Step 2: Algebraic Simplification

Simplify algebraically.Techniques:

• Factorization
• Cancellation
• Rationalization

Example:\[\frac{x^2-y^2}{x-y}\]Factor:\[x^2-y^2=(x-y)(x+y)\]Cancel:\[x+y\]

Step 3: Recognize Known Limit Patterns

Example:\[\lim_{x\to0}\frac{\sin x}{x}=1\]\[\lim_{x\to0}\frac{e^x-1}{x}=1\]

Step 4: Polar Coordinates

For limits near \((0,0)\) use\[x=r\cos\theta\]\[y=r\sin\theta\]Then evaluate\[\lim_{r\to0} f(r\cos\theta,r\sin\theta)\]If result depends on \(\theta\) → limit does not exist.

Step 5: Path Testing

Test multiple paths:

• \(x=0\)
• \(y=0\)
• \(y=mx\)

If results differ → limit does not exist.

Step 6: Mixed Term Pattern

Expressions like\[\frac{xy}{x^2+y^2}\]Use substitution\[y=mx\]If result depends on \(m\) → limit does not exist.

Step 7: Bounded Oscillation

Example:\[(x^2+y^2)\sin\left(\frac{1}{x^2+y^2}\right)\]Use the Squeeze Theorem.\[-(x^2+y^2) \le (x^2+y^2)\sin\left(\frac{1}{x^2+y^2}\right) \le (x^2+y^2)\]Both bounds → 0Therefore limit = 0.