Question

find the limit. write ∞ or - ∞ where appropriate.
lim_{x→1} \frac{-4}{(x - 1)^2}
lim_{x→1} \frac{-4}{(x - 1)^2}=\square (simplify your answer.)

Explanation:

Step1: Analyze the denominator as x approaches 1

Let \(t=(x - 1)\). As \(x
ightarrow1\), then \(t
ightarrow0\). The limit becomes \(\lim_{t
ightarrow0}\frac{- 4}{t^{2}}\).

Step2: Consider the sign and value of the denominator

When \(t
ightarrow0\), \(t^{2}>0\) for \(t
eq0\). And we have a constant - 4 in the numerator.

Step3: Determine the limit value

Since the numerator is negative (-4) and the denominator \(t^{2}\) approaches 0 from the positive - side as \(t
ightarrow0\), the value of the fraction \(\frac{-4}{t^{2}}\) approaches \(-\infty\).

Answer:

\(-\infty\)