The given expression is:

$\lim_{{x \to 2}} (2x^2 + 3x - 5)$

To find this limit, we can simply substitute $x = 2$ directly into the expression since it's a polynomial function and is continuous.

1. Substitute $x = 2$ into the expression: $2(2^2) + 3(2) - 5 = 2(4) + 6 - 5$

2. Simplify: $8 + 6 - 5 = 9$

So, the limit is:

$\lim_{{x \to 2}} (2x^2 + 3x - 5) = 9$

Would you like a detailed explanation of the steps or more examples?

Here are 5 related questions for further practice:

1. What is $\lim_{{x \to 3}} (4x^2 - 5x + 2)$?
2. Can you find $\lim_{{x \to -1}} (x^3 + 2x^2 - 4x + 3)$?
3. What happens if you have $\lim_{{x \to 2}} \frac{x^2 - 4}{x - 2}$?
4. What is the behavior of $\lim_{{x \to 0}} \frac{\sin(x)}{x}$?
5. How do you handle $\lim_{{x \to 1}} \frac{1}{x - 1}$?

Tip: For polynomials, limits at any point can be computed by direct substitution because they are continuous functions.