Find the following limits or state that they do not exist. Assume a, b, c, and k are fixed real numbers.lim⁡x→01−cos⁡(x)cos⁡2(x)−3cos⁡(x)+2{\displaystyle\lim_{x\to0}\frac{1-\cos\left(x\right)}{\$$cos^2$$\left(x\right)-3\cos\left(x\right)+2}}

Ch. 2 - Limits

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 2 - Limits

Problema 2.3.61

Capitolo 2, Problema 2.3.61

Find the following limits or state that they do not exist. Assume a, b, c, and k are fixed real numbers.

$\lim_{x \to 0} \frac{1 - \cos (x)}{\cos^{2} (x) - 3 \cos (x) + 2}$

Guida verificata passo dopo passo

Step 1: Recognize that the limit involves a trigonometric expression. The numerator is $1 - \cos(x)$, which can be approximated using the Taylor series expansion for $\cos(x)$ around $x = 0$: $\cos(x) \approx 1 - \frac{x^2}{2}$. Therefore, $1 - \cos(x) \approx \frac{x^2}{2}$ for small $x$.

Step 2: Factor the denominator $\cos^2(x) - 3\cos(x) + 2$. Notice that this is a quadratic in terms of $\cos(x)$. Let $y = \cos(x)$, then the expression becomes $y^2 - 3y + 2$. Factor this quadratic to get $(y - 1)(y - 2)$. Therefore, $\cos^2(x) - 3\cos(x) + 2 = (\cos(x) - 1)(\cos(x) - 2)$.

Step 3: Substitute the factored form of the denominator back into the limit expression: $\lim_{x \to 0} \frac{1 - \cos(x)}{(\cos(x) - 1)(\cos(x) - 2)}$. Notice that $\cos(x) - 1$ is a common factor in both the numerator and the denominator.

Step 4: Simplify the expression by canceling the common factor $\cos(x) - 1$ from the numerator and the denominator. This gives $\lim_{x \to 0} \frac{1}{\cos(x) - 2}$.

Step 5: Evaluate the simplified limit as $x$ approaches 0. Since $\cos(0) = 1$, substitute $x = 0$ into the simplified expression to find the limit.

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Limits

Limits are fundamental in calculus, representing the value that a function approaches as the input approaches a certain point. In this question, we are tasked with finding the limit of a function as x approaches 0. Understanding how to evaluate limits, especially when direct substitution leads to indeterminate forms, is crucial for solving the problem.

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