Skip to main content
Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
7장, 문제 7.5.69

Theory and Applications
L’Hôpital’s Rule does not help with the limits in Exercises 69–76. 
Try it—you just keep on cycling. Find the limits some other way.
69. lim (x → ∞) (√(9x + 1)) / (√(x + 1))

검증된 단계별 안내
1
Identify the limit expression: \(\lim_{x \to \infty} \frac{\sqrt{9x + 1}}{\sqrt{x + 1}}\).
Since both numerator and denominator involve square roots of expressions that grow without bound as \(x \to \infty\), try to simplify the expression by factoring out \(x\) inside the square roots.
Rewrite the numerator and denominator as \(\sqrt{x \left(9 + \frac{1}{x}\right)}\) and \(\sqrt{x \left(1 + \frac{1}{x}\right)}\) respectively.
Use the property of square roots to separate the factors: \(\frac{\sqrt{x} \sqrt{9 + \frac{1}{x}}}{\sqrt{x} \sqrt{1 + \frac{1}{x}}}\).
Cancel \(\sqrt{x}\) from numerator and denominator, then evaluate the limit of the remaining expression \(\frac{\sqrt{9 + \frac{1}{x}}}{\sqrt{1 + \frac{1}{x}}}\) as \(x \to \infty\) by substituting \(\frac{1}{x} \to 0\).

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
2m
도움이 되었나요?

주요 개념

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

Limits at Infinity

Limits at infinity describe the behavior of a function as the input grows without bound. Understanding how functions behave as x approaches infinity helps determine the end behavior and whether the limit converges to a finite value, infinity, or does not exist.
추천 영상:
03:07
Cases Where Limits Do Not Exist

Simplifying Expressions Involving Radicals

When dealing with limits involving square roots, it is often helpful to factor and simplify the expression inside the radicals. This can involve dividing numerator and denominator by the highest power of x to reveal dominant terms and simplify the limit evaluation.
추천 영상:
6:36
Simplifying Trig Expressions

L’Hôpital’s Rule and Its Limitations

L’Hôpital’s Rule is used to evaluate indeterminate forms like 0/0 or ∞/∞ by differentiating numerator and denominator. However, it may fail or cause cyclic behavior if repeated application does not resolve the indeterminate form, requiring alternative methods such as algebraic simplification.
추천 영상:
05:50
One-Sided Limits