IDNLearn.com is designed to help you find reliable answers to any question you have. Join our platform to receive prompt and accurate responses from experienced professionals in various fields.
Sagot :
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]
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Find reliable answers at IDNLearn.com. Thanks for stopping by, and come back for more trustworthy solutions.
