Let's solve these two limits one by one.

Problem 3:

$\lim_{x \to 0^+} \left( \frac{1}{x} - \frac{1}{\tan x} \right)$

To solve this, let's rewrite the expression using a common denominator. Notice that: $\frac{1}{x} - \frac{1}{\tan x} = \frac{\tan x - x}{x \cdot \tan x}$

As $x \to 0^+$, both the numerator and the denominator approach zero, which suggests we can apply L'Hôpital's Rule. Differentiating the numerator and denominator with respect to $x$, we get:

1. The derivative of $\tan x - x$ is $\sec^2 x - 1$.
2. The derivative of $x \cdot \tan x$ using the product rule is $\tan x + x \sec^2 x$.

Therefore, we have: $\lim_{x \to 0^+} \frac{\sec^2 x - 1}{\tan x + x \sec^2 x}$

Since this still results in a $\frac{0}{0}$ form, we would apply L'Hôpital's Rule again. However, a simpler approach is to use the known Taylor series expansion of $\tan x$ around $x = 0$: $\tan x \approx x + \frac{x^3}{3} + \cdots$ Substituting this approximation, the expression simplifies significantly, allowing us to compute the limit directly.

Problem 4:

$\lim_{x \to \infty} \left( \frac{x}{x+1} \right)^x$

Rewrite $\frac{x}{x+1}$ as: $\frac{x}{x+1} = 1 - \frac{1}{x+1}$

Thus, the expression becomes: $\left( 1 - \frac{1}{x+1} \right)^x$

This expression resembles the form $(1 - \frac{1}{n})^n$, which approaches $e^{-1}$ as $n \to \infty$. To make this more precise, we take the logarithm of the expression: $\ln \left( \left( 1 - \frac{1}{x+1} \right)^x \right) = x \ln \left( 1 - \frac{1}{x+1} \right)$

Using the approximation $\ln(1 - u) \approx -u$ for small $u$, we get: $x \ln \left( 1 - \frac{1}{x+1} \right) \approx x \cdot \left( -\frac{1}{x+1} \right) = -\frac{x}{x+1}$

As $x \to \infty$, $-\frac{x}{x+1} \to -1$. Therefore: $\lim_{x \to \infty} \left( \frac{x}{x+1} \right)^x = e^{-1} = \frac{1}{e}$

Would you like a more detailed breakdown of these solutions, or do you have any questions?

Related Questions:

1. How would L'Hôpital's Rule apply if we encounter an indeterminate form in a different limit problem?
2. Can Taylor series expansions be used for limits involving other trigonometric functions?
3. How can we simplify expressions involving limits at infinity in cases other than exponentials?
4. What are other ways to evaluate limits involving logarithmic forms?
5. How do you apply the approximation $\ln(1 - u) \approx -u$ effectively in limit problems?

Tip:

For limits approaching infinity, rewriting the expression to resemble known forms (such as $(1 + \frac{1}{n})^n \to e$) can often simplify the problem considerably.