Answer :
Sure! Let's find the limit:
[tex]\[ \lim_{{x \to 1}} \frac{x^2 + 1}{x^2 - 3x + 2} \][/tex]
### Step 1: Substitute [tex]\(x = 1\)[/tex]
First, we should substitute [tex]\(x = 1\)[/tex] into the expression to see if it yields a determinate form:
[tex]\[ \frac{1^2 + 1}{1^2 - 3(1) + 2} = \frac{1 + 1}{1 - 3 + 2} = \frac{2}{0} \][/tex]
Since the denominator becomes 0, the expression is undefined at [tex]\( x = 1 \)[/tex]. Hence, we must find the limit using other methods, since directly substituting gives an indeterminate form of [tex]\(\frac{2}{0}\)[/tex].
### Step 2: Factor the Denominator
Factor the denominator [tex]\(x^2 - 3x + 2\)[/tex]:
[tex]\[ x^2 - 3x + 2 = (x - 1)(x - 2) \][/tex]
Now rewrite the original function with the factored form:
[tex]\[ \frac{x^2 + 1}{x^2 - 3x + 2} = \frac{x^2 + 1}{(x - 1)(x - 2)} \][/tex]
### Step 3: Analyze the Behavior Around [tex]\( x = 1 \)[/tex]
Since the denominator becomes zero at [tex]\( x = 1 \)[/tex], let's analyze the behavior of the function as [tex]\( x \)[/tex] approaches 1 from the left ([tex]\( x \to 1^- \)[/tex]) and from the right ([tex]\( x \to 1^+ \)[/tex]).
#### As [tex]\( x \to 1^- \)[/tex]:
- Numerator: [tex]\( x^2 + 1 \to 1 + 1 = 2 \)[/tex]
- Denominator: [tex]\( (x - 1)(x - 2) \)[/tex] where [tex]\( x - 1 \)[/tex] approaches 0 but is slightly negative (since [tex]\( x < 1 \)[/tex]), and [tex]\( x - 2 \)[/tex] is negative when [tex]\( x < 1 \)[/tex].
So, [tex]\((x - 1)(x - 2)\)[/tex] will be positive but very close to 0.
Thus, [tex]\( \frac{2}{(x - 1)(x - 2)} \to \frac{2}{+0} \)[/tex] which indicates [tex]\( +\infty \)[/tex].
#### As [tex]\( x \to 1^+ \)[/tex]:
- Numerator: [tex]\( x^2 + 1 \to 1 + 1 = 2 \)[/tex]
- Denominator: [tex]\( (x - 1)(x - 2) \)[/tex] where [tex]\( x - 1 \)[/tex] approaches 0 but is slightly positive (since [tex]\( x > 1 \)[/tex]), and [tex]\( x - 2 \)[/tex] is negative when [tex]\( x < 2 \)[/tex].
So, [tex]\((x - 1)(x - 2)\)[/tex] will be negative but very close to 0.
Thus, [tex]\( \frac{2}{(x - 1)(x - 2)} \to \frac{2}{-0} \)[/tex] which indicates [tex]\( -\infty \)[/tex].
### Conclusion:
Since as [tex]\( x \)[/tex] approaches 1 from the left, the function approaches [tex]\( +\infty \)[/tex], and as [tex]\( x \)[/tex] approaches 1 from the right, the function approaches [tex]\( -\infty \)[/tex], the overall limit does not exist in the standard sense. However, in one-sided limits, and for certain conventions, it can be summarized as:
[tex]\[ \lim_{{x \to 1}} \frac{x^2 + 1}{x^2 - 3x + 2} = -\infty \][/tex]
Thus, the given limit evaluates to:
[tex]\[ \boxed{-\infty} \][/tex]
[tex]\[ \lim_{{x \to 1}} \frac{x^2 + 1}{x^2 - 3x + 2} \][/tex]
### Step 1: Substitute [tex]\(x = 1\)[/tex]
First, we should substitute [tex]\(x = 1\)[/tex] into the expression to see if it yields a determinate form:
[tex]\[ \frac{1^2 + 1}{1^2 - 3(1) + 2} = \frac{1 + 1}{1 - 3 + 2} = \frac{2}{0} \][/tex]
Since the denominator becomes 0, the expression is undefined at [tex]\( x = 1 \)[/tex]. Hence, we must find the limit using other methods, since directly substituting gives an indeterminate form of [tex]\(\frac{2}{0}\)[/tex].
### Step 2: Factor the Denominator
Factor the denominator [tex]\(x^2 - 3x + 2\)[/tex]:
[tex]\[ x^2 - 3x + 2 = (x - 1)(x - 2) \][/tex]
Now rewrite the original function with the factored form:
[tex]\[ \frac{x^2 + 1}{x^2 - 3x + 2} = \frac{x^2 + 1}{(x - 1)(x - 2)} \][/tex]
### Step 3: Analyze the Behavior Around [tex]\( x = 1 \)[/tex]
Since the denominator becomes zero at [tex]\( x = 1 \)[/tex], let's analyze the behavior of the function as [tex]\( x \)[/tex] approaches 1 from the left ([tex]\( x \to 1^- \)[/tex]) and from the right ([tex]\( x \to 1^+ \)[/tex]).
#### As [tex]\( x \to 1^- \)[/tex]:
- Numerator: [tex]\( x^2 + 1 \to 1 + 1 = 2 \)[/tex]
- Denominator: [tex]\( (x - 1)(x - 2) \)[/tex] where [tex]\( x - 1 \)[/tex] approaches 0 but is slightly negative (since [tex]\( x < 1 \)[/tex]), and [tex]\( x - 2 \)[/tex] is negative when [tex]\( x < 1 \)[/tex].
So, [tex]\((x - 1)(x - 2)\)[/tex] will be positive but very close to 0.
Thus, [tex]\( \frac{2}{(x - 1)(x - 2)} \to \frac{2}{+0} \)[/tex] which indicates [tex]\( +\infty \)[/tex].
#### As [tex]\( x \to 1^+ \)[/tex]:
- Numerator: [tex]\( x^2 + 1 \to 1 + 1 = 2 \)[/tex]
- Denominator: [tex]\( (x - 1)(x - 2) \)[/tex] where [tex]\( x - 1 \)[/tex] approaches 0 but is slightly positive (since [tex]\( x > 1 \)[/tex]), and [tex]\( x - 2 \)[/tex] is negative when [tex]\( x < 2 \)[/tex].
So, [tex]\((x - 1)(x - 2)\)[/tex] will be negative but very close to 0.
Thus, [tex]\( \frac{2}{(x - 1)(x - 2)} \to \frac{2}{-0} \)[/tex] which indicates [tex]\( -\infty \)[/tex].
### Conclusion:
Since as [tex]\( x \)[/tex] approaches 1 from the left, the function approaches [tex]\( +\infty \)[/tex], and as [tex]\( x \)[/tex] approaches 1 from the right, the function approaches [tex]\( -\infty \)[/tex], the overall limit does not exist in the standard sense. However, in one-sided limits, and for certain conventions, it can be summarized as:
[tex]\[ \lim_{{x \to 1}} \frac{x^2 + 1}{x^2 - 3x + 2} = -\infty \][/tex]
Thus, the given limit evaluates to:
[tex]\[ \boxed{-\infty} \][/tex]
