The image shows the limit:

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

Solving the limit:

We need to find the value of:

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

1. Direct Substitution:

   • If we substitute $x = 0$ directly, we get: $\frac{{\cos(0)}}{0} = \frac{1}{0}$ which is undefined because division by zero is not allowed.

2. Using L'Hopital's Rule: L'Hopital's Rule applies when we have an indeterminate form like $\frac{0}{0}$. Here, we already know direct substitution leads to undefined, so we proceed with L'Hopital's Rule:

   • First, differentiate the numerator and denominator: $\text{Numerator: } \frac{d}{dx} [\cos(x)] = -\sin(x)$ $\text{Denominator: } \frac{d}{dx} [x] = 1$
   • Apply L'Hopital's Rule: $\lim_{{x \to 0}} \frac{{\cos(x)}}{x} = \lim_{{x \to 0}} \frac{{-\sin(x)}}{1}$

3. Substitute $x = 0$ again: $\lim_{{x \to 0}} (-\sin(x)) = -\sin(0) = 0$

Thus, the final answer is:

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

Would you like any further details or have any questions?

Here are 5 related questions to expand on this:

1. What are the conditions for applying L'Hopital's Rule?
2. How do we differentiate $\sin(x)$ and $\cos(x)$?
3. What happens to $\lim_{{x \to 0}} \frac{\sin(x)}{x}$?
4. Can L'Hopital's Rule be applied to limits approaching infinity?
5. What other methods, besides L'Hopital's Rule, can be used to evaluate limits involving trigonometric functions?

Tip: Always check if a limit can be simplified or solved by substitution before applying more complex methods like L'Hopital's Rule.