Q.$\lim_{x\to 1}\dfrac{ax^2 + bx + c}{cx^2 + bx + a},\ a + b + c \neq 0$

The key idea is that for a polynomial limit where direct substitution gives the indeterminate form $\frac{0}{0}$, we factor out the common root $(x-1)$ from numerator and denominator. Here, since $a+b+c=0$ is not given, direct substitution is valid.

Step 1: Substitute $x = 1$ directly into the expression:

The key idea is that when you have a limit of a polynomial divided by another polynomial, and the denominator doesn't vanish at the point you're approaching, you can simply plug in the value. This works because polynomials are continuous functions — their graphs have no jumps or holes at any real number.

Here, both the numerator $ax^2 + bx + c$ and the denominator $cx^2 + bx + a$ are polynomials in $x$. At $x=1$, the denominator becomes $c(1)^2 + b(1) + a = a + b + c$. The problem explicitly tells us $a + b + c \neq 0$, so the denominator is non-zero at $x=1$. That means the rational function is continuous at $x=1$, and the limit equals the function's value there.

Let's walk through it step by step.

1. Identify the form of the limit.

   We have $\lim_{x\to 1} \frac{ax^2 + bx + c}{cx^2 + bx + a}$. Both numerator and denominator are quadratic polynomials. There's no $\frac{0}{0}$ or $\frac{\infty}{\infty}$ situation here because the denominator at $x=1$ is $a+b+c$, which is given to be non-zero.

2. Apply direct substitution.

   Since the denominator is non-zero at $x=1$, the limit is simply the value of the fraction at $x=1$:

3. Simplify the fraction. Because $a+b+c \neq 0$, we can cancel the common factor: $$$$