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Ch. 2 - Limits
Briggs - Calculus: Early Transcendentals 3rd Edition
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2장, 문제 45b

Analyze the following limits and find the vertical asymptotes of f(x) = (x − 5) / (x2 − 25).
lim x → -5- f(x)

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Step 1: Identify the points where the function f(x) = \( \frac{x - 5}{x^2 - 25} \) is undefined. This occurs when the denominator is zero. Set \( x^2 - 25 = 0 \) and solve for x.
Step 2: Factor the denominator \( x^2 - 25 \) as a difference of squares: \( (x - 5)(x + 5) = 0 \). This gives the points x = 5 and x = -5 where the function is undefined.
Step 3: Determine if these points are vertical asymptotes by checking the behavior of the function as x approaches these values. For x = -5, consider the limit \( \lim_{x \to -5^-} \frac{x - 5}{(x - 5)(x + 5)} \).
Step 4: Simplify the expression by canceling the common factor (x - 5) in the numerator and denominator, resulting in \( \lim_{x \to -5^-} \frac{1}{x + 5} \).
Step 5: Evaluate the limit \( \lim_{x \to -5^-} \frac{1}{x + 5} \). As x approaches -5 from the left, the denominator approaches 0 from the negative side, indicating a vertical asymptote at x = -5.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Limits

Limits describe the behavior of a function as the input approaches a certain value. In this context, we are interested in the limit of f(x) as x approaches -5 from the left, which helps determine the function's behavior near that point. Understanding limits is crucial for analyzing continuity and identifying asymptotic behavior.
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Vertical Asymptotes

Vertical asymptotes occur at values of x where a function approaches infinity or negative infinity, typically where the denominator of a rational function equals zero while the numerator does not. For the function f(x) = (x − 5) / (x² − 25), we need to find the values of x that make the denominator zero to identify potential vertical asymptotes.
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Factoring Polynomials

Factoring polynomials involves rewriting a polynomial as a product of its factors, which can simplify expressions and help identify roots. In this case, the denominator x² - 25 can be factored into (x - 5)(x + 5), allowing us to easily find the points where the function is undefined and analyze the limits around those points.
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