Understanding Algebraic Limits: Techniques and Examples

Algebraic Limits

Algebraic limits are a concept in calculus that involves finding the limiting value of a function as it approaches a certain value or as it approaches infinity or negative infinity

Algebraic limits are a concept in calculus that involves finding the limiting value of a function as it approaches a certain value or as it approaches infinity or negative infinity. These limits are often used to evaluate the behavior of functions and determine their asymptotic behavior.

To find algebraic limits, you can use several techniques such as factoring, rationalizing, or applying algebraic manipulations. Let’s go through a few examples to understand how algebraic limits work.

Example 1:
Find the limit of f(x) = (3x^2 + 2x + 1) / (5x^2 – 4x + 3) as x approaches 2.
To find the limit, substitute the value of x into the function:
f(2) = (3(2)^2 + 2(2) + 1) / (5(2)^2 – 4(2) + 3)
f(2) = (12 + 4 + 1) / (20 – 8 + 3)
f(2) = 17 / 15
So, the limit of f(x) as x approaches 2 is 17/15.

Example 2:
Find the limit of g(x) = (x^2 – 4) / (x – 2) as x approaches 2.
In this example, if we directly substitute x = 2, we get an indeterminate form of 0/0. So, let’s factor the numerator:
g(x) = (x – 2)(x + 2) / (x – 2)
Now, we can cancel out the common factor of (x – 2):
g(x) = (x + 2)
Now substitute x = 2:
g(2) = 2 + 2
g(2) = 4
Therefore, the limit of g(x) as x approaches 2 is 4.

Example 3:
Find the limit of h(x) = (sqrt(x + 1) – 1) / (x – 2) as x approaches 2.
Let’s rationalize the numerator to simplify the expression:
h(x) = [(sqrt(x + 1) – 1) * (sqrt(x + 1) + 1)] / (x – 2) * (sqrt(x + 1) + 1)
h(x) = (x + 1 – 1) / (x – 2) * (sqrt(x + 1) + 1)
h(x) = x / (x – 2) * (sqrt(x + 1) + 1)
Now, we can substitute x = 2:
h(2) = 2 / (2 – 2) * (sqrt(2 + 1) + 1)
We get an indeterminate form of 0/0. To evaluate this limit further, we can simplify the expression as follows:
h(2) = 2 / 0 * (sqrt(3) + 1)
Since we have a 0 in the denominator, this indicates that the function has a vertical asymptote at x = 2. Therefore, the limit of h(x) as x approaches 2 does not exist.

These examples illustrate different scenarios in algebraic limits. Remember that direct substitution is not always sufficient, and you may need to apply algebraic manipulations to simplify the expression and evaluate the limit.

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