Let's analyze and solve the limit problems shown in the image one by one.

a. $\lim_{x \to 0} \frac{\sin(x)}{x}$

This is a well-known limit and is typically solved using standard limit properties. The result is:

$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$

b. $\lim_{x \to 0} \frac{x}{\cos(x)}$

This limit can be computed directly by substitution since $\cos(0) = 1$. Therefore:

$\lim_{x \to 0} \frac{x}{\cos(x)} = \frac{0}{1} = 0$

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

We can use L'Hopital's Rule here since the limit is in the indeterminate form $\frac{0}{0}$.

• Differentiate the numerator: $\frac{d}{dx}[(x-1)e^x] = e^x + (x-1)e^x$
• Differentiate the denominator: $\frac{d}{dx}[\ln(x)] = \frac{1}{x}$

Applying L'Hopital's Rule:

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

d. $\lim_{x \to 0^+} x \ln(x)$

This limit can be solved by noting that as $x \to 0^+$, $\ln(x) \to -\infty$ and $x \to 0^+$, so the product tends to $0 \times (-\infty)$, which is indeterminate.

To solve this, we can rewrite $x \ln(x)$ as $\frac{\ln(x)}{1/x}$, and then apply L'Hopital's Rule.

• Differentiate the numerator: $\frac{d}{dx}[\ln(x)] = \frac{1}{x}$
• Differentiate the denominator: $\frac{d}{dx}[1/x] = -\frac{1}{x^2}$

Now the limit becomes:

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

Thus:

$\lim_{x \to 0^+} x \ln(x) = 0$

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

Here we have a limit in the form of $1^\infty$, which is indeterminate. To evaluate this, we take the natural logarithm and use L'Hopital's Rule.

First, rewrite the expression as:

$y = (\cos(x))^{\frac{1}{x^2}} \quad \Rightarrow \quad \ln(y) = \frac{1}{x^2} \ln(\cos(x))$

Now, we find:

$\lim_{x \to 0} \ln(y) = \lim_{x \to 0} \frac{\ln(\cos(x))}{x^2}$

Apply L'Hopital's Rule since this is in the indeterminate form $\frac{0}{0}$:

• Differentiate the numerator: $\frac{d}{dx}[\ln(\cos(x))] = -\frac{\sin(x)}{\cos(x)} = -\tan(x)$
• Differentiate the denominator: $\frac{d}{dx}[x^2] = 2x$

Now the limit becomes:

$\lim_{x \to 0} \frac{-\tan(x)}{2x} = \lim_{x \to 0} \frac{-\sin(x)}{2x\cos(x)} = \frac{-1}{2}$

So:

$\lim_{x \to 0} \ln(y) = -\frac{1}{2}$

Exponentiate both sides to get $y$:

$y = e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}}$

Thus:

$\lim_{x \to 0} (\cos(x))^{\frac{1}{x^2}} = \frac{1}{\sqrt{e}}$

Let me know if you need any further details or have any questions! Here are some additional problems related to limits:

1. What is the limit of $\frac{\sin(2x)}{x}$ as $x \to 0$?
2. How would you solve $\lim_{x \to \infty} \frac{e^x}{x^2}$?
3. Can you evaluate $\lim_{x \to 0} \frac{e^x - 1}{x}$?
4. What is $\lim_{x \to 0} \frac{\ln(1+x)}{x}$?
5. How do you find $\lim_{x \to \infty} \frac{2x^3 + x}{5x^3 - 3x}$?

Tip: Always check if a limit leads to an indeterminate form like $\frac{0}{0}$ or $\infty/\infty$ to determine if you should use L'Hopital's Rule.