Thank you for sharing the image! I see a set of limit problems in the photo. Let me analyze and solve them one by one. I'll go through each step in detail for clarity.

1. $\lim_{x \to 0} \frac{x^3 - 3x^2 + 2}{x^3 - 4x^2 + 3}$

This is a rational function. As $x \to 0$: $\text{Substitute } x = 0: \frac{(0)^3 - 3(0)^2 + 2}{(0)^3 - 4(0)^2 + 3} = \frac{2}{3}.$ Solution: $\frac{2}{3}$.

2. $\lim_{x \to 0} \frac{e^x - e^{-x}}{\ln(1 + x)}$

Expand using Taylor series:

• $e^x \approx 1 + x + \frac{x^2}{2} + \dots$,
• $e^{-x} \approx 1 - x + \frac{x^2}{2} + \dots$,
• $\ln(1+x) \approx x - \frac{x^2}{2} + \dots$.

Simplify numerator: $e^x - e^{-x} \approx (1 + x + \frac{x^2}{2}) - (1 - x + \frac{x^2}{2}) = 2x.$

Simplify denominator: $\ln(1+x) \approx x.$

Thus: $\frac{e^x - e^{-x}}{\ln(1+x)} \approx \frac{2x}{x} = 2.$ Solution: $2$.

3. $\lim_{x \to \infty} \frac{\pi - 2\arctan(x)}{e^{\frac{x}{2}} - 1}$

As $x \to \infty$:

• $\arctan(x) \to \frac{\pi}{2}$,
• $e^{\frac{x}{2}} \to \infty$.

For the numerator: $\pi - 2\arctan(x) \to \pi - 2\left(\frac{\pi}{2}\right) = 0.$

For the denominator: $e^{\frac{x}{2}} - 1 \to \infty.$

Thus: $\frac{\pi - 2\arctan(x)}{e^{\frac{x}{2}} - 1} \to \frac{0}{\infty} = 0.$ Solution: $0$.

4. $\lim_{x \to 0} \frac{2 - (e^x + e^{-x})\cos(x)}{x^4}$

Expand the terms using Taylor series:

• $e^x \approx 1 + x + \frac{x^2}{2} + \dots$,
• $e^{-x} \approx 1 - x + \frac{x^2}{2} + \dots$,
• $\cos(x) \approx 1 - \frac{x^2}{2} + \dots$.

Simplify the numerator: $e^x + e^{-x} \approx (1 + x + \frac{x^2}{2}) + (1 - x + \frac{x^2}{2}) = 2 + x^2.$ $(e^x + e^{-x})\cos(x) \approx (2 + x^2)\left(1 - \frac{x^2}{2}\right) \approx 2 - x^2 + x^2 = 2.$ $2 - (e^x + e^{-x})\cos(x) \approx 0.$

Since the numerator approaches $0$ faster than $x^4$, the limit is $0$. Solution: $0$.

5. $\lim_{x \to 0} \frac{e^{3x} - 3x - 1}{\sin^2(5x)}$

Expand the numerator using Taylor series: $e^{3x} \approx 1 + 3x + \frac{(3x)^2}{2} = 1 + 3x + \frac{9x^2}{2}.$ $e^{3x} - 3x - 1 \approx (1 + 3x + \frac{9x^2}{2}) - 3x - 1 = \frac{9x^2}{2}.$

For the denominator: $\sin^2(5x) \approx (5x)^2 = 25x^2.$

Thus: $\frac{e^{3x} - 3x - 1}{\sin^2(5x)} \approx \frac{\frac{9x^2}{2}}{25x^2} = \frac{9}{50}.$ Solution: $\frac{9}{50}$.

6. $\lim_{x \to 0} \frac{\sin(3x) - 3x e^x + 3x^2}{\arctan(x) - \sin(x) - \frac{x^3}{6}}$

Numerator:

• $\sin(3x) \approx 3x - \frac{(3x)^3}{6} = 3x - \frac{27x^3}{6}$,
• $3x e^x \approx 3x(1 + x + \frac{x^2}{2}) = 3x + 3x^2 + \frac{3x^3}{2}$,
• $3x^2$ remains as is.

$\sin(3x) - 3x e^x + 3x^2 \approx (3x - \frac{27x^3}{6}) - (3x + 3x^2 + \frac{3x^3}{2}) + 3x^2.$ $= -3x^2 - \frac{27x^3}{6} - \frac{3x^3}{2} = -3x^2 - \frac{45x^3}{6}.$

Denominator:

• $\arctan(x) \approx x - \frac{x^3}{3}$,
• $\sin(x) \approx x - \frac{x^3}{6}$,
• Combine with $-\frac{x^3}{6}$: $\arctan(x) - \sin(x) - \frac{x^3}{6} \approx (x - \frac{x^3}{3}) - (x - \frac{x^3}{6}) - \frac{x^3}{6}.$ $= -\frac{x^3}{3} + \frac{x^3}{6} - \frac{x^3}{6} = -\frac{x^3}{3}.$

Simplify:

$\frac{-3x^2 - \frac{45x^3}{6}}{-\frac{x^3}{3}} \approx \frac{-3x^2}{-\frac{x^3}{3}} = \frac{9}{x}.$ [ = \ldots\text{ Will write the continuation. Fixing! **Solution Is...