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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 Savage Hardy. in South Africa
Mar 14,2025

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Answer

The limits are: a) 0 b) 0 c) 7/5 d) 1/4

Solution

**a) Evaluate \( \lim_{x \rightarrow 0} \frac{1-\cos (x)}{2 x} \):** 1. Use the Taylor series expansion for \(\cos(x)\) around \(x=0\): \[ \cos(x) = 1 - \frac{x^2}{2} + O(x^4) \] 2. Thus, the numerator becomes: \[ 1 - \cos(x) \approx \frac{x^2}{2} \] 3. Substitute into the limit: \[ \frac{1-\cos(x)}{2x} \approx \frac{\frac{x^2}{2}}{2x} = \frac{x^2}{4x} = \frac{x}{4} \] 4. Taking the limit as \(x \to 0\): \[ \lim_{x \rightarrow 0} \frac{x}{4} = 0 \] --- **b) Evaluate \( \lim_{x \rightarrow 0} \frac{1-\cos (x)}{\sin (x)} \):** 1. Again, using the Taylor series expansion: \[ 1 - \cos(x) \approx \frac{x^2}{2} \] and note that: \[ \sin(x) \approx x \] 2. Substitute the approximations into the limit: \[ \frac{1-\cos(x)}{\sin(x)} \approx \frac{\frac{x^2}{2}}{x} = \frac{x}{2} \] 3. Taking the limit as \(x \to 0\): \[ \lim_{x \rightarrow 0} \frac{x}{2} = 0 \] --- **c) Evaluate \( \lim_{x \rightarrow 0} \frac{\sin (7x)}{\sin (5x)} \):** 1. Use the small-angle approximation: \[ \sin(7x) \approx 7x \quad \text{and} \quad \sin(5x) \approx 5x \] 2. Substitute into the limit: \[ \frac{\sin(7x)}{\sin(5x)} \approx \frac{7x}{5x} = \frac{7}{5} \] 3. Therefore: \[ \lim_{x \rightarrow 0} \frac{\sin(7x)}{\sin(5x)} = \frac{7}{5} \] --- **d) Evaluate \( \lim_{x \rightarrow 2} \frac{\sin (x-2)}{x^{2}-4} \):** 1. Factor the denominator: \[ x^2 - 4 = (x-2)(x+2) \] 2. Rewrite the limit: \[ \frac{\sin(x-2)}{x^2-4} = \frac{\sin(x-2)}{(x-2)(x+2)} \] 3. Split the limit into two parts: \[ \lim_{x \rightarrow 2} \frac{\sin(x-2)}{x-2} \cdot \lim_{x \rightarrow 2} \frac{1}{x+2} \] 4. Recognize that: \[ \lim_{x \rightarrow 2} \frac{\sin(x-2)}{x-2} = 1 \quad \text{(by the standard limit } \lim_{u \rightarrow 0} \frac{\sin u}{u}=1 \text{)} \] 5. And: \[ \lim_{x \rightarrow 2} \frac{1}{x+2} = \frac{1}{2+2} = \frac{1}{4} \] 6. Multiply the results: \[ 1 \cdot \frac{1}{4} = \frac{1}{4} \] --- **Final Answers:** a) \(0\) b) \(0\) c) \(\frac{7}{5}\) d) \(\frac{1}{4}\)

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Let's evaluate each of the limits one by one. ### a) \[ \lim _{x \rightarrow 0} \frac{1-\cos (x)}{2 x} \] Using the identity \(1 - \cos(x) = 2 \sin^2(\frac{x}{2})\), we can rewrite the limit: \[ \frac{1 - \cos(x)}{2x} = \frac{2 \sin^2(\frac{x}{2})}{2x} = \frac{\sin^2(\frac{x}{2})}{x} \] As \(x\) approaches 0, \(\frac{x}{2} \to 0\) as well, so we can use the small-angle approximation \(\sin(u) \approx u\): \[ = \lim_{x \to 0} \frac{\sin^2(\frac{x}{2})}{x} = \lim_{u \to 0} \frac{u^2}{2u} = \lim_{u \to 0} \frac{u}{2} = 0 \] Thus: \[ \text{Result: } 0 \] --- ### b) \[ \lim _{x \rightarrow 0} \frac{1-\cos (x)}{\sin (x)} \] Again, using the identity \(1 - \cos(x) = 2 \sin^2(\frac{x}{2})\): \[ \frac{1 - \cos(x)}{\sin(x)} = \frac{2 \sin^2(\frac{x}{2})}{\sin(x)} \] Using \(\sin(x) = 2 \sin(\frac{x}{2}) \cos(\frac{x}{2})\): \[ \frac{2 \sin^2(\frac{x}{2})}{2 \sin(\frac{x}{2}) \cos(\frac{x}{2})} = \frac{\sin(\frac{x}{2})}{\cos(\frac{x}{2})} \] As \(x\) approaches 0, \(\frac{x}{2} \to 0\) which leads to: \[ \frac{\frac{x}{2}}{1} = 0 \] Thus: \[ \text{Result: } 0 \] --- ### c) \[ \lim _{x \rightarrow 0} \frac{\sin (7 x)}{\sin (5 x)} \] Using L'Hôpital's Rule since both the numerator and denominator approach 0: \[ = \lim_{x \to 0} \frac{7 \cos(7x)}{5 \cos(5x)} = \frac{7\cos(0)}{5\cos(0)} = \frac{7}{5} \] Thus: \[ \text{Result: } \frac{7}{5} \] --- ### d) \[ \lim _{x \rightarrow 2} \frac{\sin (x-2)}{x^{2}-4} \] We can rewrite the denominator as \(x^2 - 4 = (x - 2)(x + 2)\): \[ \lim _{x \to 2} \frac{\sin(x-2)}{(x - 2)(x + 2)} \] As \(x \to 2\): \[ = \lim_{x \to 2} \frac{\sin(x-2)}{x-2} \cdot \frac{1}{x + 2} \] From the small-angle limit, \(\frac{\sin(u)}{u} \to 1\) as \(u \to 0\): \[ = 1 \cdot \frac{1}{4} = \frac{1}{4} \] Thus: \[ \text{Result: } \frac{1}{4} \] --- In summary, the limits are: a) 0, b) 0, c) \(\frac{7}{5}\), d) \(\frac{1}{4}\).

Related Questions

If you invest \( \$ 1000 \) for \( t \) years at \( 6.71 \% \) compounded at the frequencies below, find the following. (a) Suppose you compound at \( 6.71 \% \) monthly. i) Report an expression equivalent to the value of \( \$ 1000 \) invested for \( t \) years at \( 6.71 \% \) compounded monthly by completing the box with the growth factor if compounded annually. 1000 \( \square \) Number \( t \) (Round to \( \underline{4} \) decimal places.) ii) Report the effective annual rate: \( \square \) Number \% (Round to \( \underline{2} \) decimal places.) (b) Suppose you compound at \( 6.71 \% \) continuously. i) You would expect \( 6.71 \% \) compounded continuously to give a \( \square \) Click for List yield than what is given in part (a). ii) Complete the boxes below to report the expression for the value of \( \$ 1000 \) invested for \( t \) years at \( 6.71 \% \) compounded continuously and the equivalent growth factor if compounded annually. \[ \begin{array}{l} 1000 e^{(\text {Number } t)} \\ \approx 1000(\text { Number })^{t} \end{array} \] (Round to \( \underline{4} \) decimal places.) iii) Report the effective annual rate: \( \square \) Number \% (Round to \( \underline{2} \) decimal places.) (c) Complete the boxes to summarize: i) From part (a) we have that 6.71 \% compounded monthly is equivalent to \( \square \) Number \( \% \) compounded annually. ii) From part (b) we have that 6.71 \% compounded continuously is equivalent to \( \square \) Number \( \% \) compounded annually.

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