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Magazine Chapter 2: Problem 33
Find the limits in Exercises \(19-36\). $$ \lim _{x \rightarrow 2} \frac{\sqrt{x^{2}+12}-4}{x-2} $$
Short Answer
Expert verified
The limit is \( \frac{1}{2} \).
Step by step solution
01
Identify Type of Limit
Examine the expression \( \frac{\sqrt{x^{2}+12}-4}{x-2} \) as \( x \rightarrow 2 \). It involves a square root and subtraction in the numerator, suggesting a need for algebraic manipulation to simplify.
02
Rationalize the Numerator
To eliminate the square root from the numerator, multiply the numerator and the denominator by the conjugate \( \sqrt{x^{2}+12}+4 \) to rationalize it. This gives us:\[\frac{\left[\sqrt{x^{2}+12}-4\right]\left[\sqrt{x^{2}+12}+4\right]}{(x-2)(\sqrt{x^{2}+12}+4)}\]
03
Simplify the Expression
Multiply the expressions in the numerator using the difference of squares:\[(\sqrt{x^{2}+12})^2 - 4^2 = x^2 + 12 - 16 = x^2 - 4\]The expression now is:\[\frac{x^2 - 4}{(x-2)(\sqrt{x^{2}+12}+4)}\]
04
Factor the Quadratic Expression
Recognize that \( x^2 - 4 \) is a difference of squares and can be factored as \((x-2)(x+2)\). Substitute to obtain:\[\frac{(x-2)(x+2)}{(x-2)(\sqrt{x^{2}+12}+4)}\]
05
Cancel the Common Factors
Cancel the common factor \( x-2 \) from both the numerator and the denominator:\[\frac{x+2}{\sqrt{x^{2}+12}+4}\]
06
Evaluate the Limit
Substitute \( x = 2 \) into the simplified expression:\[\lim_{x \to 2} \frac{x+2}{\sqrt{x^{2}+12}+4} = \frac{2+2}{\sqrt{2^2+12}+4} = \frac{4}{\sqrt{16}+4} = \frac{4}{8} = \frac{1}{2}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rationalization
When dealing with limits involving square roots, rationalization is a useful technique. This process helps in eliminating the square root from the numerator, which simplifies the expression. In the exercise, the numerator is \( \sqrt{x^{2}+12} - 4 \). To rationalize it, we multiply both the numerator and the denominator by the conjugate \( \sqrt{x^{2}+12} + 4 \).
This step transforms the original limit:
This step transforms the original limit:
- Before: \( \frac{\sqrt{x^{2}+12} - 4}{x-2} \)
- After multiplying by the conjugate: \( \frac{\left[\sqrt{x^{2}+12} - 4\right]\left[\sqrt{x^{2}+12} + 4\right]}{(x-2)(\sqrt{x^{2}+12}+4)} \)
Difference of Squares
The difference of squares is a useful algebraic identity that simplifies expressions into products. It's expressed generally as \( a^2 - b^2 = (a-b)(a+b) \). In this limit problem, rationalization leads to a difference of squares in the numerator, \( (\sqrt{x^{2}+12})^2 - 4^2 \).
By applying the difference of squares concept:
By applying the difference of squares concept:
- \( (\sqrt{x^{2}+12})^2 = x^2 + 12 \)
- \( 4^2 = 16 \)
- So, \( x^2 + 12 - 16 = x^2 - 4 \)
Factoring Polynomials
Factoring is another essential algebraic tool, especially when simplifying expressions in calculus. The polynomial \( x^2 - 4 \) can be factored using another special algebraic identity, the difference of squares. This is because \( x^2 - 4 \) can be rewritten as \((x-2)(x+2)\).
Steps:
Steps:
- Recognize \( x^2 - 4 \) as a difference of squares.
- Factor it: \( x^2 - 4 = (x-2)(x+2) \).
Evaluating Limits
The final step is to evaluate the limit. After simplifying the expression through rationalization and factoring, we have the limits expression as \( \frac{x+2}{\sqrt{x^{2}+12}+4} \).
Since we successfully cancelled the problematic \( x-2 \) factor, we can substitute \( x = 2 \) directly without worrying about division by zero:
This method ensures an accurate and meaningful understanding of how limits involving complex fractions are evaluated neatly and precisely.
Since we successfully cancelled the problematic \( x-2 \) factor, we can substitute \( x = 2 \) directly without worrying about division by zero:
- \( x+2 \) becomes \( 2+2 = 4 \).
- \( \sqrt{x^{2}+12}+4 \) becomes \( \sqrt{2^2+12} + 4 = \sqrt{16} + 4 = 4 + 4 = 8 \).
This method ensures an accurate and meaningful understanding of how limits involving complex fractions are evaluated neatly and precisely.
