Pregunta
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7. Evaluate the following limits: a) \( \lim _{x \rightarrow 0} \frac{1-\cos (x)}{2 x} \) b) \( \lim _{x \rightarrow 0} \frac{1-\cos (x)}{\sin (x)} \) c) \( \lim _{x \rightarrow 0} \frac{\sin (7 x)}{\sin (5 x)} \) d) \( \lim _{x \rightarrow 2} \frac{\sin (x-2)}{x^{2}-4} \).

Ask by Paul Garza. in South Africa
Mar 14,2025

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**a)** 0 **b)** 0 **c)** 7/5 **d)** 1/4

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Beyond the Answer

Let's evaluate each of the limits step by step: a) To evaluate \( \lim _{x \rightarrow 0} \frac{1-\cos (x)}{2 x} \), we can use the Taylor expansion of \( \cos(x) \) near \( x=0 \): \[ \cos(x) \approx 1 - \frac{x^2}{2} + O(x^4). \] Thus, \[ 1 - \cos(x) \approx \frac{x^2}{2}. \] Substituting this into our limit gives: \[ \lim_{x \rightarrow 0} \frac{1 - \cos(x)}{2x} \approx \lim_{x \rightarrow 0} \frac{\frac{x^2}{2}}{2x} = \lim_{x \rightarrow 0} \frac{x}{4} = 0. \] b) For the limit \( \lim _{x \rightarrow 0} \frac{1-\cos (x)}{\sin (x)} \), again employing the Taylor expansion: \[ 1 - \cos(x) \approx \frac{x^2}{2}, \quad \text{and} \quad \sin(x) \approx x. \] So we rewrite the limit: \[ \lim_{x \rightarrow 0} \frac{\frac{x^2}{2}}{x} = \lim_{x \rightarrow 0} \frac{x}{2} = 0. \] c) Now for \( \lim _{x \rightarrow 0} \frac{\sin (7 x)}{\sin (5 x)} \), we use the small-angle approximation \( \sin(kx) \approx kx \): \[ \lim_{x \rightarrow 0} \frac{\sin(7x)}{\sin(5x)} \approx \lim_{x \rightarrow 0} \frac{7x}{5x} = \frac{7}{5}. \] d) Finally, for \( \lim _{x \rightarrow 2} \frac{\sin (x-2)}{x^{2}-4} \), notice that both the numerator and the denominator approach \(0\) as \(x \to 2\). Use L'Hôpital's Rule: \[ \frac{d}{dx}[\sin(x - 2)] = \cos(x - 2), \quad \text{and} \quad \frac{d}{dx}[x^2 - 4] = 2x. \] Now apply L'Hôpital's Rule: \[ \lim_{x \rightarrow 2} \frac{\cos(x - 2)}{2x} = \frac{\cos(0)}{2(2)} = \frac{1}{4}. \] In summary, the limits are: a) \( 0 \) b) \( 0 \) c) \( \frac{7}{5} \) d) \( \frac{1}{4} \)

preguntas relacionadas

Step 1 We are asked to find the surface area of the curve defined by \( y=x^{3} \) rotated about the \( x \)-axis over the interval \( 0 \leq x \leq 2 \). Recall the following formula for the surface area of a function of \( x \) rotated about the \( x \)-axis. Note that as the curve rotates in a circular manner about the \( x \)-axis, the expression \( 2 \pi y \) is the circumference of radius \( y \) and the radical measures the arc length that is the width of a band. \[ S=\int_{a}^{b} 2 \pi y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x \] We begin by substituting \( y=x^{3} \) and its derivative in the surface area formula and simplifying. \[ \begin{aligned} S & \left.=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+\left(\sqrt{3 x^{2}}\right.} \sqrt{3 x^{2}}\right)^{2} d x \\ & =\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9}, 9^{x^{4}} d x \end{aligned} \] Step 2 We have found the following integral for the surface area. \[ S=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9 x^{4}} d x \] To evaluate the integral we will first make the substitution \( u=1+9 x^{4} \). We also will need the following to complete the substitution. \[ \begin{array}{l} d u=36 x^{3} \\ x=0 \rightarrow u=1 \\ x=2 \rightarrow u=\square 146 x^{d x} \end{array} \] Step 3 We can now make the substitution \( u=1+(9 x)^{4} \) and evaluate the definite integral with respect to \( u \). \[ \begin{aligned} \int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+(9 x)^{4}} d x & =\int_{1}^{145} 2 \pi \sqrt{u}\left(\frac{d u}{36}\right) \\ & =\frac{2 \pi}{36} \int_{1}^{145} \sqrt{u} d u \end{aligned} \] \[ =\frac{2 \pi}{36}\left[\frac{2}{3} u^{\frac{2}{3}}\right]_{1}^{145} \]

Latest Calculus Questions

\[ y=x^{3}, \quad 0 \leq x \leq 2 \] Step 1 We are asked to find the surface area of the curve defined by \( y=x^{3} \) rotated about the \( x \)-axis over the interv \( 0 \leq x \leq 2 \). Recall the following formula for the surface area of a function of \( x \) rotated about the \( x \)-axis. Note t as the curve rotates in a circular manner about the \( x \)-axis, the expression \( 2 \pi y \) is the circumference of radius and the radical measures the arc length that is the width of a band, \[ S=\int_{a}^{b} 2 \pi y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x \] We begin by substituting \( y=x^{3} \) and its derivative in the surface area formula and simplifying, \[ \begin{aligned} S & \left.=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+\left(\sqrt{3 x^{2}}\right.} \sqrt{3 x^{2}}\right)^{2} d x \\ & =\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9} \sqrt[9]{ } x^{4} d x \end{aligned} \] Step 2 We have found the following integral for the surface area. \[ S=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9 x^{4}} d x \] To evaluate the integral we will first make the substitution \( u=1+9 x^{4} \). We also will need the following to complete the substitution. \[ \begin{array}{l} d u=36 x^{3} \\ x=0 \rightarrow u=1 \\ x=2 \rightarrow u=\square 14 \end{array} \] Step 3 We can now make the substitution \( u=1+(9 x)^{4} \) and evaluate the definite integral with respect to \( u \). \[ \begin{aligned} \int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+(9 x)^{4}} d x & =\int_{1}^{145} 2 \pi \sqrt{u}\left(\frac{d u}{36}\right) \\ & =\frac{2 \pi}{36} \int_{1}^{145} \sqrt{u} d u \end{aligned} \] \[ =\frac{2 \pi}{36}\left[\frac{2}{3} u^{\frac{2}{3}}\right]_{1}^{145} \]
Step 1 We are asked to find the surface area of the curve defined by \( y=x^{3} \) rotated about the \( x \)-axis over the interval \( 0 \leq x \leq 2 \). Recall the following formula for the surface area of a function of \( x \) rotated about the \( x \)-axis. Note that as the curve rotates in a circular manner about the \( x \)-axis, the expression \( 2 \pi y \) is the circumference of radius \( y \) and the radical measures the arc length that is the width of a band. \[ S=\int_{a}^{b} 2 \pi y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x \] We begin by substituting \( y=x^{3} \) and its derivative in the surface area formula and simplifying. \[ \begin{aligned} S & \left.=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+\left(\sqrt{3 x^{2}}\right.} \sqrt{3 x^{2}}\right)^{2} d x \\ & =\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9}, 9^{x^{4}} d x \end{aligned} \] Step 2 We have found the following integral for the surface area. \[ S=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9 x^{4}} d x \] To evaluate the integral we will first make the substitution \( u=1+9 x^{4} \). We also will need the following to complete the substitution. \[ \begin{array}{l} d u=36 x^{3} \\ x=0 \rightarrow u=1 \\ x=2 \rightarrow u=\square 146 x^{d x} \end{array} \] Step 3 We can now make the substitution \( u=1+(9 x)^{4} \) and evaluate the definite integral with respect to \( u \). \[ \begin{aligned} \int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+(9 x)^{4}} d x & =\int_{1}^{145} 2 \pi \sqrt{u}\left(\frac{d u}{36}\right) \\ & =\frac{2 \pi}{36} \int_{1}^{145} \sqrt{u} d u \end{aligned} \] \[ =\frac{2 \pi}{36}\left[\frac{2}{3} u^{\frac{2}{3}}\right]_{1}^{145} \]
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