שחזור ופתרון אפשרי - בוחן אמצע

At first glance I was mistaken by \( sin(\frac{\pi}{2}) = 1 \) - I assumed the function is continuous and \( lim_{x \to \frac{\pi}{4}} \frac{1}{sin(2x)} = \frac{1}{sin(\frac{\pi}{2})} = 1 \)

However, if you place \( x = \frac{\pi}{4} \) in the original function, you get \( f(x) = \frac{\frac{f(x) + g(x)}{\underbrace{f(x) - g(x)}}_{=0} + \frac{f(x) - g(x)}{f(x) + g(x)}}{\frac{f(x) + g(x)}{\underbrace{f(x) - g(x)}_{=0}} - \frac{f(x) - g(x)}{f(x) + g(x)}} \), since at \( x = \frac{\pi}{4} \), we have: \[ f(x) - g(x) = sin(\frac{\pi}{4}) - cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0 \] So the function is not continuous at \( x = \frac{\pi}{4} \)

Answer: c) has a removable discontinuity

Explanation:

1. For simplicity let \( t = \frac{1}{x} \), so the limit becomes: \[ \lim_{t \to 4} \frac{t - 4}{\frac{1}{t} - \frac{1}{4}} \]
2. We can simplify the expression: \[ \lim_{t \to 4} \frac{t - 4}{\frac{4 - t}{4t}} \]
3. So the limit becomes: \[ \lim_{t \to 4} \frac{4t \cdot (t - 4)}{(-1) \cdot (t - 4)} = \lim_{t \to 4} -4t = -16 \text{ (since } t = \frac{1}{x} \text{)} \]

We can also use the sandwich theorem: \[ \frac{-4}{x} \leq \frac{\frac{1}{x} - 4}{x - \frac{1}{4}} \leq \frac{\frac{1}{x}}{- \frac{1}{4}} \] Since \( \lim_{x \to \frac{1}{4}} \frac{-4}{x} = -16 \) and \( \lim_{x \to \frac{1}{4}} \frac{\frac{1}{x}}{- \frac{1}{4}} = -16 \), the limit is -16 as well.

Answer: a) \((-3, 1)\)

Explanation:

We can simplify the function to:

\[ f(x) = \begin{cases} \frac{\sin(2x)}{x(x^2 + 2x - 3)} & \text{if } x \neq 0 \\ - \frac{2}{3} & \text{if } x = 0 \end{cases} \]

I dont rember the question well but I did notice that:

• The function is continuous in the open interval (-3,1), but not in the closed interval [-3,1]
• The function is not defined at x = -3, 1
• The function is continuous at x = 0

Answer: c) \( \lim_{x \to \pi / 2} \sin(x) ^ \frac{\pi}{cos^2(x)} = e^{-\pi/2} \)

Explanation:

The expression is similar to \( 1^\infty \), so we can use the limit \( \lim_{x \to 0} (1 + x)^\frac{1}{x} = e \)

To do so, notice that \( \sin x = \left( 1 - {\cos^2 x} \right) ^ {1/2} \) since \( \sin^2 x = 1 - \cos^2 x \)

So the limit becomes: \[ \lim_{x \to \pi / 2} \left[\left( 1 - {\cos^2 x} \right) ^{\frac{1}{2}} \right]^{\frac{\pi}{\cos^2 x}} \]

Rearranging the expression, we get: \[ \lim_{x \to \pi / 2} \left[\left( 1 - {\cos^2 x} \right) ^{\frac{1}{ \cos^2 x}}\right]^{\frac{\pi}{2}} \]

We can take out \( \frac{\pi}{2} \) from the limit, since a constant factor doesn't affect the limit, and we get: \[ \left[ \lim_{x \to \pi / 2} \left( 1 - {\cos^2 x} \right) ^{\frac{1}{ \cos^2 x}} \right]^{\frac{\pi}{2}} \]

We can replace \( -\cos^2 x \) with \( y = -\cos^2 x \), so the limit becomes: \[ \left[ \lim_{y \to 0} \left( 1 + y \right) ^{\frac{1}{y}} \right]^{-\left(\frac{\pi}{2}\right)} = \boxed{e^{-\pi/2}} \]

We can declare: \[ y = \left(\sin x\right)^{\frac{\pi}{\cos^2 x}} \Rightarrow \ln y = \frac{\pi}{\cos^2 x} \ln \left( \sin x\right)\] And we get: \[ \lim_{x \to \pi / 2} \left( \ln y \right) = \pi \lim_{x \to \pi / 2} \left( \frac{\ln \left( \sin x\right)}{\cos^2 x}\right) \]

This expression is from the form \( \frac{0}{0} \), so we can use L'Hopital's Rule: \[ \pi \lim_{x \to \pi / 2} \left( \frac{\ln \left( \sin x\right)}{\cos^2 x}\right) = \pi \lim_{x \to \pi / 2} \left( \frac{\frac{ \cancel{ \cos x }}{\sin x}}{-2 \cancel{ \cos x} \sin x}\right) = \pi \lim_{x \to \pi / 2} \left( \frac{-1}{2 \sin^2 x}\right) = - \frac{\pi}{2} \]

So we get: \[ \ln y = - \frac{\pi}{2} \Rightarrow y = e^{-\pi/2} \]

1. Let's analyze this step by step. We have: - f, g are continuous at point x₀ - h has limit L at x₀
2. Looking at the innermost function: \[\lim_{x \to x_0} h(x) = L\]
3. Since g is continuous at x₀, then: \[\lim_{x \to x_0} g^{h(x)} = g^{\lim_{x \to x_0} h(x)} = g^L\]
4. Finally, since f is continuous, we can apply it: \[\lim_{x \to x_0} f\left(g^{h(x)}\right) = f\left(\lim_{x \to x_0} g^{h\left(x\right)}\right) = f(g^L)\]

Answer: a) f(g^L)