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Consider the function \(f(x) = \dfrac{x-1}{x^2-1}\). We want to determine its limit as \(x \to 1\).
Justify that direct substitution leads to an indeterminate form.
Show that for \(x \neq 1\), \(\dfrac{x-1}{x^2-1} = \dfrac{1}{x+1}\).
Conclude about the value of \(\displaystyle\lim_{x \to 1} \dfrac{x-1}{x^2-1}\).
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