Example Question - evaluating limits
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Evaluating Limits in Calculus
<p>Чтобы решить данный предел, воспользуемся правилом Лопиталя, так как при подстановке \( x = 0 \) получается неопределенность вида \( \frac{0}{0} \).</p> <p>Исходная функция: \( \lim_{x \to 0} \frac{\sin x - x}{x^3} \)</p> <p>Применим правило Лопиталя и возьмем производные числителя и знаменателя:</p> <p>Производная числителя: \( (\sin x - x)' = \cos x - 1 \)</p> <p>Производная знаменателя: \( (x^3)' = 3x^2 \)</p> <p>Применяем правило Лопиталя еще раз, так как получившаяся дробь снова приводит к неопределенности \( \frac{0}{0} \), при \( x = 0 \):</p> <p>Производная числителя после первого применения Лопиталя: \( (\cos x - 1)' = -\sin x \)</p> <p>Производная знаменателя после первого применения Лопиталя: \( (3x^2)' = 6x \)</p> <p>Таким образом, получаем:</p> <p>\( \lim_{x \to 0} \frac{-\sin x}{6x} \)</p> <p>Мы можем упростить это выражение, зная, что \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \):</p> <p>\( \lim_{x \to 0} \frac{-\sin x}{6x} = \lim_{x \to 0} \frac{-1}{6} \cdot \frac{\sin x}{x} = \frac{-1}{6} \cdot 1 = -\frac{1}{6} \)</p> <p>Итак, ответ: \( -\frac{1}{6} \).</p>
Evaluating Limits of Functions
The question displays three functions and asks to evaluate the limits of certain expressions involving these functions. Given: - \( f: R \rightarrow R \) with \( f(x) = x^2 + 5x + 4 \) - \( g: R \rightarrow R \) with \( g(x) = 3x - 3 \) We are asked to find the following limits: a) \(\lim_{x \to -2} [f(x) + g(x)] \) b) \(\lim_{x \to 2} \sqrt{f(x)} - g(x) \) c) \(\lim_{x \to 3} \frac{f(x)}{g(x)} \) Let's solve each one: a) To find the limit of \( f(x) + g(x) \) as \( x \) approaches \(-2\), we simply plug \( -2 \) into each function and add the results: \( f(-2) = (-2)^2 + 5(-2) + 4 = 4 - 10 + 4 = -2 \) \( g(-2) = 3(-2) - 3 = -6 - 3 = -9 \) Now, \( f(-2) + g(-2) = -2 + (-9) = -11 \) So, \(\lim_{x \to -2} [f(x) + g(x)] = -11\). b) To find the limit of \( \sqrt{f(x)} - g(x)\) as \(x\) approaches \(2\), we substitute \(2\) into each function and then perform the operations, keeping in mind to take the square root of \(f(x)\) not \(f(x)\) itself: \( f(2) = (2)^2 + 5(2) + 4 = 4 + 10 + 4 = 18 \) \( g(2) = 3(2) - 3 = 6 - 3 = 3 \) Now, the square root of \( f(2) \) is \( \sqrt{18} \), so: \( \sqrt{f(2)} - g(2) = \sqrt{18} - 3 \) This result cannot be simplified further without a calculator, so we leave it as \( \sqrt{18} - 3 \). Therefore, \(\lim_{x \to 2} \sqrt{f(x)} - g(x) = \sqrt{18} - 3\). c) To find the limit of \( \frac{f(x)}{g(x)}\) as \(x\) approaches \(3\), we again substitute \(3\) into each function and divide the results: \( f(3) = (3)^2 + 5(3) + 4 = 9 + 15 + 4 = 28 \) \( g(3) = 3(3) - 3 = 9 - 3 = 6 \) Now, \( \frac{f(3)}{g(3)} = \frac{28}{6} = \frac{14}{3} \) So, \(\lim_{x \to 3} \frac{f(x)}{g(x)} = \frac{14}{3}\). To summarize, the limits are as follows: a) \(-11\) b) \( \sqrt{18} - 3 \) c) \( \frac{14}{3} \)
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