To find the limit

$\lim_{x \to 2} \frac{x+2}{(x-2)^2}$

we can start by analyzing the behavior of the numerator and the denominator as $x \to 2$.

1. Substitute $x = 2$:
   • In the numerator, $x + 2 = 2 + 2 = 4$.
   • In the denominator, $(x - 2)^2 = (2 - 2)^2 = 0$.

This gives us a form of $\frac{4}{0}$, which indicates a division by zero, suggesting that the limit may approach infinity or negative infinity. We need to examine the direction of approach for $x \to 2$.

2. Analyze the sign as $x \to 2^+$ and $x \to 2^-$:
   • When $x \to 2^+$ (approaching from the right), $x - 2$ is a small positive number, so $(x - 2)^2$ is a small positive value.
   • When $x \to 2^-$ (approaching from the left), $x - 2$ is a small negative number, but $(x - 2)^2$ is still a small positive value (since it’s squared).

In both cases, the denominator approaches $0^+$, a small positive value, while the numerator remains close to $4$. Therefore:

$\lim_{x \to 2} \frac{x+2}{(x-2)^2} = +\infty$

because the fraction becomes very large as $x$ gets closer to 2 from either side.

Final Answer

$\lim_{x \to 2} \frac{x+2}{(x-2)^2} = +\infty$

Would you like more details on this solution, or do you have further questions?

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1. What is an example of a limit approaching infinity with a different denominator?
2. How does the square in $(x - 2)^2$ affect the behavior of this limit?
3. How would the result change if the numerator were $x - 2$ instead?
4. What happens if you take a similar limit but with $(x - 2)^3$ in the denominator?
5. Can limits approach negative infinity? Under what conditions?

Tip: When encountering a zero in the denominator, consider approaching from both sides to see if the limit goes to positive or negative infinity.