The expression you're asking to evaluate is:

$\lim_{x \to 1} \left( \frac{1}{x} - \frac{1}{x - 1} \right)$

At $x = 1$, both terms will cause problems:

• The term $\frac{1}{x}$ will become $\frac{1}{1} = 1$.
• The term $\frac{1}{x - 1}$ will result in a division by zero.

Because of the second term, we need to proceed by simplifying the expression. Let’s first combine the two fractions over a common denominator:

$\frac{1}{x} - \frac{1}{x - 1} = \frac{(x - 1) - x}{x(x - 1)} = \frac{-1}{x(x - 1)}$

Now, the expression becomes:

$\lim_{x \to 1} \frac{-1}{x(x - 1)}$

As $x \to 1$, the denominator $x(x - 1)$ approaches zero. Specifically, $x - 1$ approaches 0 from either side, so the entire expression tends to infinity (or negative infinity, since the numerator is negative).

Thus, the limit does not exist in a traditional sense, as the expression tends to either positive or negative infinity depending on the direction of approach. Therefore, we say:

$\lim_{x \to 1} \left( \frac{1}{x} - \frac{1}{x - 1} \right) = -\infty$

Do you want more details or have any questions about this?

Here are 5 related questions you might find interesting:

1. How do limits behave at points of discontinuity?
2. Can you apply L'Hôpital's Rule to solve limits involving indeterminate forms?
3. How do you compute limits involving trigonometric functions?
4. What is the significance of one-sided limits in calculus?
5. How can limits help in understanding asymptotic behavior of functions?

Tip: When you encounter a limit that involves division by zero, consider whether the limit tends to infinity or whether it can be resolved using algebraic simplifications or special techniques like L'Hôpital's Rule.