Example Question - l'hôpital's rule
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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 the Limit of a Trigonometric Function
\[ \lim_{{x \to 0}} \frac{{\sin(x) - x}}{{x^3}} = \lim_{{x \to 0}} \frac{{\sin(x) - x}}{{x^3}} \cdot \frac{{\sin(x) + x}}{{\sin(x) + x}} = \lim_{{x \to 0}} \frac{{\sin^2(x) - x^2}}{{x^3(\sin(x) + x)}} \] \[ = \lim_{{x \to 0}} \frac{{\sin^2(x) - x^2}}{{x^3}} \cdot \lim_{{x \to 0}} \frac{1}{{\sin(x) + x}} \] \[ = \lim_{{x \to 0}} \frac{{(\sin(x) + x)(\sin(x) - x)}}{{x^3}} \cdot \lim_{{x \to 0}} \frac{1}{{\sin(x) + x}} \] Using the identity $(a+b)(a-b) = a^2 - b^2$ and L'Hôpital's Rule, \[ = \lim_{{x \to 0}} \frac{{\sin(x) - x}}{{x^2}} \cdot \lim_{{x \to 0}} \frac{1}{{\sin(x) + x}} \] Applying L'Hôpital's Rule for the first limit, \[ = \lim_{{x \to 0}} \frac{{\cos(x) - 1}}{{2x}} \] \[ = \frac{1}{2} \cdot \lim_{{x \to 0}} \frac{{\cos(x) - 1}}{{x}} \] Applying L'Hôpital's Rule once more, \[ = \frac{1}{2} \cdot \lim_{{x \to 0}} \frac{{-\sin(x)}}{1} \] \[ = \frac{1}{2} \cdot \lim_{{x \to 0}} -\sin(x) \] \[ = \frac{1}{2} \cdot 0 \] \[ = 0 \]
Solving Limit using Trigonometric Identity and L'Hôpital's Rule
Para resolver el límite \( \lim_{x \to 0} \frac{1 - \cos(ax)}{1 - \cos(bx)} \), podemos aplicar la identidad trigonométrica \( \cos(\theta) = 1 - 2\sin^2(\frac{\theta}{2}) \) para transformar tanto el numerador como el denominador. Así, la expresión se convierte en: \[ \lim_{x \to 0} \frac{1 - (1 - 2\sin^2(\frac{ax}{2}))}{1 - (1 - 2\sin^2(\frac{bx}{2}))} \] Simplificando el numerador y el denominador, obtenemos: \[ \lim_{x \to 0} \frac{2\sin^2(\frac{ax}{2})}{2\sin^2(\frac{bx}{2})} \] Podemos cancelar el factor 2 en el numerador y en el denominador, lo que nos deja: \[ \lim_{x \to 0} \frac{\sin^2(\frac{ax}{2})}{\sin^2(\frac{bx}{2})} \] Ahora podemos aplicar el límite usando la regla de L'Hôpital, ya que tenemos una forma indeterminada de tipo 0/0. Pero antes de aplicar la regla de L'Hôpital, vamos a simplificar aún más tomando la raíz cuadrada al numerador y al denominador, recordando que la raíz cuadrada es una función continua y que podemos tomar el límite fuera de ella: \[ \lim_{x \to 0} \left(\frac{\sin(\frac{ax}{2})}{\sin(\frac{bx}{2})}\right)^2 = \left(\lim_{x \to 0} \frac{\sin(\frac{ax}{2})}{\sin(\frac{bx}{2})}\right)^2 \] Utilizando la regla de L'Hôpital diferenciamos el numerador y el denominador por separado: \[ \frac{d}{dx}\sin\left(\frac{ax}{2}\right) = \frac{a}{2}\cos\left(\frac{ax}{2}\right) \] \[ \frac{d}{dx}\sin\left(\frac{bx}{2}\right) = \frac{b}{2}\cos\left(\frac{bx}{2}\right) \] Ahora, aplicando la derivada a la expresión original: \[ \lim_{x \to 0} \frac{\frac{a}{2}\cos\left(\frac{ax}{2}\right)}{\frac{b}{2}\cos\left(\frac{bx}{2}\right)} = \lim_{x \to 0} \frac{a \cos\left(\frac{ax}{2}\right)}{b \cos\left(\frac{bx}{2}\right)} \] Dado que \( \cos(0) = 1 \), la expresión se simplifica a: \[ \frac{a}{b} \] Por lo tanto, el límite original es \( \left(\frac{a}{b}\right)^2 \), que corresponde a la opción (c) en tu lista de opciones si hubiera una. (La imagen no muestra ninguna opción después de la letra "b").
Analyzing Limit of a Rational Function as x Approaches 3
The image contains a handwritten math problem asking to find the limit of a rational function as x tends to 3. The function is given by: f(x) = (x^4 - 8x^3 + 22x + 2) / (x^5 + 3x^2 - 6x + 3) To find the limit as x approaches 3, we would typically substitute the value of 3 into the function. However, this function seems to be indeterminate at x = 3 since the denominator becomes zero. We would need to do some algebraic manipulation to simplify the expression and eliminate the indeterminate form. Given the complexity of the polynomial, a common approach would be to factor both the numerator and the denominator, possibly by finding common factors or applying polynomial division. Unfortunately, the higher degree of these polynomials makes factoring directly a non-trivial process without additional information or context. A potential approach here is to factor by grouping or to use synthetic division to check if (x - 3) is a factor of the numerator and denominator, but the process is not shown in the image. Another approach could be to apply L'Hôpital's Rule, which states that if the limit yields an indeterminate form of 0/0 or ∞/∞, we can take the derivative of the numerator and the derivative of the denominator and then find the limit of that new function as x approaches the same value. However, given the degree and complexity of the polynomials involved, this could be quite laborious without further simplification. Without additional context or simplification methods provided in the image, I am unable to give a complete solution to the problem.
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