Infinite Limits — Question 9

Question 9 - Solution

We are asked to evaluate: $$\lim_{x \to 0^-} \frac{2x + 1}{x}$$

Step 1: Behavior near zero from the left

As $x \to 0^-$, the numerator $2x + 1 \to 1$, since $2x \to 0^-$ and $1$ is a constant.

The denominator $x \to 0^-$, which is a small negative number.

So the entire fraction behaves like: $$\frac{2x + 1}{x} \approx \frac{1}{\text{tiny negative}} = -\infty$$

Step 2: More formal analysis (optional)

We can write: $$\frac{2x + 1}{x} = 2 + \frac{1}{x}$$

As $x \to 0^-$, $\frac{1}{x} \to -\infty$, and the constant 2 becomes negligible in comparison.

Conclusion: $$\boxed{\lim_{x \to 0^-} \frac{2x + 1}{x} = -\infty}$$