1. **Problem Statement:** Find the slope of the secant line PQ where $P=(1,0)$ and $Q=(x, \sin(\frac{10\pi}{x}))$ for given values of $x$. Then analyze if these slopes approach a limit and estimate the slope of the tangent line at $P$. 2. **Formula for slope of secant line:** The slope $m_{PQ}$ of the secant line through points $P$ and $Q$ is given by: $$ m_{PQ} = \frac{y_Q - y_P}{x - 1} = \frac{\sin\left(\frac{10\pi}{x}\right) - 0}{x - 1} = \frac{\sin\left(\frac{10\pi}{x}\right)}{x - 1} $$ 3. **Calculate slopes for given $x$ values:** Calculate $m_{PQ}$ for $x = 2, 1.5, 1.4, 1.3, 1.2, 1.1, 0.5, 0.6, 0.7, 0.8, 0.9$. - For $x=2$: $$m_{PQ} = \frac{\sin\left(\frac{10\pi}{2}\right)}{2-1} = \frac{\sin(5\pi)}{1} = 0$$ - For $x=1.5$: $$m_{PQ} = \frac{\sin\left(\frac{10\pi}{1.5}\right)}{0.5} = \frac{\sin\left(\frac{20\pi}{3}\right)}{0.5} = \frac{\sin\left(2\pi + \frac{2\pi}{3}\right)}{0.5} = \frac{\sin\left(\frac{2\pi}{3}\right)}{0.5} = \frac{\sqrt{3}/2}{0.5} = 1.7321$$ - For $x=1.4$: $$m_{PQ} = \frac{\sin\left(\frac{10\pi}{1.4}\right)}{0.4} \approx \frac{\sin(22.448)}{0.4} \approx \frac{0.9749}{0.4} = 2.4373$$ - For $x=1.3$: $$m_{PQ} = \frac{\sin\left(\frac{10\pi}{1.3}\right)}{0.3} \approx \frac{\sin(24.132)}{0.3} \approx \frac{-0.4067}{0.3} = -1.3557$$ - For $x=1.2$: $$m_{PQ} = \frac{\sin\left(\frac{10\pi}{1.2}\right)}{0.2} \approx \frac{\sin(26.179)}{0.2} \approx \frac{-0.9945}{0.2} = -4.9725$$ - For $x=1.1$: $$m_{PQ} = \frac{\sin\left(\frac{10\pi}{1.1}\right)}{0.1} \approx \frac{\sin(28.571)}{0.1} \approx \frac{0.4339}{0.1} = 4.3390$$ - For $x=0.5$: $$m_{PQ} = \frac{\sin\left(\frac{10\pi}{0.5}\right)}{-0.5} = \frac{\sin(20\pi)}{-0.5} = \frac{0}{-0.5} = 0$$ - For $x=0.6$: $$m_{PQ} = \frac{\sin\left(\frac{10\pi}{0.6}\right)}{-0.4} \approx \frac{\sin(52.36)}{-0.4} \approx \frac{-0.8660}{-0.4} = 2.1650$$ - For $x=0.7$: $$m_{PQ} = \frac{\sin\left(\frac{10\pi}{0.7}\right)}{-0.3} \approx \frac{\sin(44.88)}{-0.3} \approx \frac{0.9749}{-0.3} = -3.2497$$ - For $x=0.8$: $$m_{PQ} = \frac{\sin\left(\frac{10\pi}{0.8}\right)}{-0.2} \approx \frac{\sin(39.27)}{-0.2} \approx \frac{-0.7071}{-0.2} = 3.5355$$ - For $x=0.9$: $$m_{PQ} = \frac{\sin\left(\frac{10\pi}{0.9}\right)}{-0.1} \approx \frac{\sin(34.90)}{-0.1} \approx \frac{-0.8660}{-0.1} = 8.6603$$ 4. **Do slopes approach a limit?** The slopes oscillate widely and do not settle near a single value as $x$ approaches 1 from either side. This is due to the highly oscillatory nature of $\sin\left(\frac{10\pi}{x}\right)$ near $x=1$. 5. **Why slopes differ from tangent slope at $P$:** The tangent slope at $P$ is the limit of secant slopes as $x \to 1$. However, because the function oscillates rapidly near $x=1$, the secant slopes do not approach a single limit, so the tangent slope is not well-defined or is very sensitive. 6. **Estimate slope of tangent line at $P$:** By choosing secant points $x$ very close to 1 and averaging or observing trends, the slope appears to oscillate between positive and negative large values, suggesting the tangent slope at $P$ does not exist or is undefined. **Final answer:** The slopes of secant lines do not approach a finite limit near $x=1$, so the slope of the tangent line at $P$ is not well-defined.