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@@ -199,15 +199,20 @@ $$\lim_{x \to -\infty} \frac{1}{x^2}$$
Similar to the previous case, as x approaches negative infinity, the denominator ($x^2$) grows infinitely large (but remains positive), and the numerator remains constant at 1.
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The fraction becomes increasingly small, approaching zero.
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Solution:
$$\lim_{x \to -\infty} \frac{1}{x^2} = 0$$
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