Find the value of each limit: the limit of 25 as x approaches 1, of 3x^2 as x approaches 10, of 2x^4 - x^2 as x approaches 3, of (x^3 + 2)^6 as x approaches 1, and of |-15x + 4| as x approaches 3.

Step-by-step solution

1. 1

   Check for continuity before doing anything else. Direct substitution is valid exactly when the function is continuous at the target point. Polynomials are continuous everywhere, absolute values of polynomials are continuous everywhere, and constants are trivially continuous — so all five limits here are simply function values. There is no $0/0$ to resolve.

2. 2

   Limit 1: a constant function. $f(x) = 25$ never changes, so approaching $x = 1$ changes nothing:

   $\lim_{x\to 1} 25 = 25$

   The presence of $x$ in the limit notation does not mean $x$ appears in the function.

3. 3

   Limits 2 and 3: polynomials. Substitute and evaluate carefully, respecting the exponents:

   $\lim_{x\to 10} 3x^2 = 3(10)^2 = 3(100) = 300$

   $\lim_{x\to 3} (2x^4 - x^2) = 2(81) - 9 = 162 - 9 = 153$

   Note $3(10)^2 = 300$, not $(3\cdot 10)^2 = 900$ — the coefficient is not squared.

4. 4

   Limit 4: a power of a polynomial. Evaluate the inside first, then raise:

   $\lim_{x\to 1} (x^3+2)^6 = (1+2)^6 = 3^6 = 729$

   This uses the composition law for limits, valid because the outer function $u^6$ is continuous.

5. 5

   Limit 5: an absolute value. Evaluate the inside, then take the absolute value:

   $\lim_{x\to 3} |-15x+4| = |-45+4| = |-41| = 41$

   The result is $+41$: an absolute value is never negative. Absolute value functions have corners, but a corner does not break continuity, so substitution is still legitimate — and in any case the corner is at $x = \tfrac{4}{15}$, nowhere near $x = 3$.

6. 6

   Collect and verify. The five values are $25$, $300$, $153$, $729$ and $41$. Each was reconfirmed with a symbolic limit engine, and each equals the plain function value at the target point, as continuity guarantees.