Animated Solution for Mathematics - Limits, Continuity and Differentiability: If ${\lim }_{x\to 1}\frac{x^2-ax+b}{x-1}=5$, then $a+b$ is equal to :-

We are tasked with evaluating the limit:

As $x\to 1$, the denominator $x-1$ approaches $0$. For the limit to exist as a finite value ($5$), the expression must represent an indeterminate form of the type $\frac{0}{0}$.

This implies that the numerator must also vanish at $x=1$. Therefore, we must satisfy the condition:

From the condition above, we derive the relationship:

We substitute this expression for $b$ back into the original numerator:

To simplify, we rearrange the terms to group the squares and the linear components:

Using the difference of squares identity, $x^2-1=(x-1)(x+1)$, we rewrite the numerator as:

Factoring out the common term $(x-1)$, we obtain:

Now, we substitute this back into the limit expression:

Since $x\to 1$ implies $x eq 1$, we can safely cancel the $(x-1)$ terms to remove the singularity.

With the indeterminate form resolved, we perform direct substitution:

Using our previously established relationship $b=a-1$, we find:

The final requirement is to calculate the sum $a+b$:

The final result is $-7$.