https://file.aiquickdraw.com/mathbot/file/1728306687101vtocs3ne.jpg

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.

Find the following limits: 
a. lim(x->0) sin(x)/x 
b. lim(x->0) x/cos(x) 
c. lim(x->1) (x-1)e^x / ln(x) 
d. lim(x->0+) x * ln(x) 
e. lim(x->0) (cos(x))^(1/x^2)

This page provides step-by-step solutions for solving limit problems involving trigonometric functions, logarithmic expressions, and the use of L'Hopital's Rule. Learn how to compute limits like lim(x->0) sin(x)/x and lim(x->1) (x-1)e^x / ln(x).

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