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Magazine Chapter 9: Problem 22
\(\operatorname{Lt}_{x \rightarrow 3} \frac{x^{-5}-\frac{1}{243}}{x-3}=?\) (1) \(-\frac{5}{729}\) (2) \(-\frac{5}{243}\) (3) \(\frac{5}{81}\) (4) None of these
Short Answer
Expert verified
Answer: (3) \(-\frac{5}{81}\)
Step by step solution
01
Simplify the expression.
Simplify the given expression by finding a common denominator for the numerator:
\(\frac{x^{-5}-\frac{1}{243}}{x-3} = \frac{243x^{-5} - 1}{243(x-3)}\)
02
Rewrite using positive exponents.
Rewrite the expression with positive exponents by letting \(y = x - 3\):
\(\frac{243(x - 3 + 3)^{-5} - 1}{243(x-3)} = \frac{243((y+3)^{-5}) - 1}{243y}\)
03
Apply the Binomial Theorem.
In this step, we apply the Binomial Theorem to simplify the numerator:
\((y+3)^{-5} = \sum_{k=0}^{\infty} \binom{-5}{k} y^{k} 3^{k-5}\)
04
Consider the first two terms.
Since we only need to find the limit as \(y\) approaches 0, we only need to consider the first two terms of the expansion:
\((y+3)^{-5} \approx \binom{-5}{0} y^{0} 3^{-5} + \binom{-5}{1} y^{1} 3^{-4}\)
05
Calculate the coefficients.
By calculating the binomial coefficients, we get:
\(\binom{-5}{0} = 1\) and \(\binom{-5}{1} = -5\)
Hence, the expression becomes:
\((y+3)^{-5} \approx 1 \cdot 3^{-5} - 5 \cdot y \cdot 3^{-4}\)
06
Substitute in the original expression.
Substitute the approximation back into the original expression:
\(\frac{243((y+3)^{-5}) - 1}{243y} \approx \frac{243((1 \cdot 3^{-5}) - 5 \cdot y \cdot 3^{-4}) - 243}{243y}\)
07
Simplify the expression.
Now, simplify the expression and find the limit as \(y\) approaches 0:
\(\lim_{y\rightarrow 0} \frac{243(1 \cdot 3^{-5}) - 5 \cdot y \cdot 3^{-4} - 243}{243y} \approx \lim_{y\rightarrow 0} \frac{-\frac{5}{81} \cdot y}{y} = -\frac{5}{81}\)
Thus, the correct answer is (3) \(\frac{5}{81}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Binomial Theorem
The Binomial Theorem is a mathematical principle that provides a method to expand expressions that are raised to a power. It is especially useful in Calculus when dealing with limits involving exponents. The theorem states, for any integer \[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]In this context, the theorem is applied to the term \((y+3)^{-5}\),which involves a negative exponent. The expansion is infinite when dealing with negative exponents, as in \((y+3)^{-5} = \sum_{k=0}^{\infty} \binom{-5}{k} y^k 3^{k-5}\).This approach helps us focus on the first few terms that will significantly influence the limit as \(y\) approaches 0.
Exponents and Powers
Exponents are a way of expressing repeated multiplication of a number by itself. For example, \(x^{-5}\) means \(1/x^5\).Now, negative exponents indicate the reciprocal of the base raised to the absolute value of the exponent. It turns an exponentiation operation into a division one, offering powerful simplification options in Calculus problems.
- Positive exponents indicate standard multiplication.
- Negative exponents mean division, or a multiplicative inverse.
Limit Calculation Steps
Calculating limits involves a process that often requires expression simplification. In this exercise, we dealt with the expression \(\operatorname{Lt}_{x \rightarrow 3} \frac{x^{-5}-\frac{1}{243}}{x-3}\).The steps to solve it include:
- Simplify: Combine terms and look for common denominators as needed.
- Rewrite: Use substitutions like \(y = x - 3\)to simplify calculations in terms of \(y\).
- Apply Theorems: Here, the Binomial Theorem helps handle expressions with exponents like\((y+3)^{-5}\).
- Approximate: Consider only significant terms (like the first two from the binomial expansion) to focus on the behavior near the limit point.
- Solve: Evaluate the limit, as the expression becomes simpler with new terms canceled out.
