Answer :
To find the limit [tex]\(\lim_{x \to 1} \frac{x^2 - 3}{(x - 1)^2}\)[/tex], we follow these steps.
1. Direct Substitution:
First, try directly substituting [tex]\( x \)[/tex] with 1 in the expression [tex]\(\frac{x^2 - 3}{(x - 1)^2}\)[/tex]:
- The numerator becomes [tex]\( 1^2 - 3 = 1 - 3 = -2 \)[/tex].
- The denominator becomes [tex]\( (1 - 1)^2 = 0^2 = 0 \)[/tex].
Thus, direct substitution leads to the form [tex]\(\frac{-2}{0}\)[/tex], which suggests that the limit may tend toward infinity or negative infinity. This form indicates that we are dealing with an infinite limit. However, we need deeper analysis to determine the exact behavior.
2. Analyzing the behavior near [tex]\( x = 1 \)[/tex]:
We can inspect the behavior of the function as [tex]\( x \)[/tex] approaches 1 from both sides:
- As [tex]\( x \to 1^+ \)[/tex] (from the right),
- The numerator [tex]\( x^2 - 3 \)[/tex] will be slightly negative (approaching -2).
- The denominator [tex]\( (x - 1)^2 \)[/tex] will be a very small positive number.
- Hence, the overall expression [tex]\(\frac{x^2 - 3}{(x - 1)^2}\)[/tex] will approach negative values divided by very small positive values, leading to [tex]\(-\infty\)[/tex].
- As [tex]\( x \to 1^- \)[/tex] (from the left),
- The numerator [tex]\( x^2 - 3 \)[/tex] will still be slightly negative (approaching -2).
- The denominator [tex]\( (x - 1)^2 \)[/tex] will also be a very small positive number since squaring removes the sign of [tex]\( x - 1 \)[/tex].
- Consequently, the overall expression [tex]\(\frac{x^2 - 3}{(x - 1)^2}\)[/tex] will likewise approach negative values divided by very small positive values, again leading to [tex]\(-\infty\)[/tex].
3. Conclusion:
Since the function approaches [tex]\(-\infty\)[/tex] as [tex]\( x \)[/tex] approaches 1 from both the left and the right, we conclude that:
[tex]\[ \lim_{x \to 1} \frac{x^2 - 3}{(x - 1)^2} = -\infty \][/tex]
This is our final answer.
1. Direct Substitution:
First, try directly substituting [tex]\( x \)[/tex] with 1 in the expression [tex]\(\frac{x^2 - 3}{(x - 1)^2}\)[/tex]:
- The numerator becomes [tex]\( 1^2 - 3 = 1 - 3 = -2 \)[/tex].
- The denominator becomes [tex]\( (1 - 1)^2 = 0^2 = 0 \)[/tex].
Thus, direct substitution leads to the form [tex]\(\frac{-2}{0}\)[/tex], which suggests that the limit may tend toward infinity or negative infinity. This form indicates that we are dealing with an infinite limit. However, we need deeper analysis to determine the exact behavior.
2. Analyzing the behavior near [tex]\( x = 1 \)[/tex]:
We can inspect the behavior of the function as [tex]\( x \)[/tex] approaches 1 from both sides:
- As [tex]\( x \to 1^+ \)[/tex] (from the right),
- The numerator [tex]\( x^2 - 3 \)[/tex] will be slightly negative (approaching -2).
- The denominator [tex]\( (x - 1)^2 \)[/tex] will be a very small positive number.
- Hence, the overall expression [tex]\(\frac{x^2 - 3}{(x - 1)^2}\)[/tex] will approach negative values divided by very small positive values, leading to [tex]\(-\infty\)[/tex].
- As [tex]\( x \to 1^- \)[/tex] (from the left),
- The numerator [tex]\( x^2 - 3 \)[/tex] will still be slightly negative (approaching -2).
- The denominator [tex]\( (x - 1)^2 \)[/tex] will also be a very small positive number since squaring removes the sign of [tex]\( x - 1 \)[/tex].
- Consequently, the overall expression [tex]\(\frac{x^2 - 3}{(x - 1)^2}\)[/tex] will likewise approach negative values divided by very small positive values, again leading to [tex]\(-\infty\)[/tex].
3. Conclusion:
Since the function approaches [tex]\(-\infty\)[/tex] as [tex]\( x \)[/tex] approaches 1 from both the left and the right, we conclude that:
[tex]\[ \lim_{x \to 1} \frac{x^2 - 3}{(x - 1)^2} = -\infty \][/tex]
This is our final answer.