Evaluate $\lim_{x\to k}\frac{x^n+ax+b}{x-k}$, where $k$ is a roots of $x^n+ax+b=0$

Question

If $a_1,a_2,..,a_n$ are the roots of the equation $x^n+ax+b=0$ then find $(a_1-a_2)(a_1-a_3)....(a_1-a_n)$

My Attempt

$$ x^n+ax+b=(x-a_1)(x-a_2)(x-a_3)...(x-a_n)=0\\ (x-a_2)(x-a_3)...(x-a_n)=\frac{x^n+ax+b}{x-a_1}\\ (a_1-a_2)(a_1-a_3)....(a_1-a_n)=\lim_{x\to a_1}(x-a_2)(x-a_3)...(x-a_n)\\ =\lim_{x\to a_1}\frac{x^n+ax+b}{x-a_1} $$ My reference gives the solution $na_1^{n-1}+a$. I can use L'Hospital's rule $$ \lim_{x\to a_1}\frac{x^n+ax+b}{x-a_1}=\lim_{x\to a_1}\frac{nx^{n-1}+a}{1}=na_1^{n-1}+a $$

But how do I evaluate the limit directly ?

Answer 1

For any differentiable function $f(x)$, if $r$ is a root of $f$, then

$$\lim_{x\to r}\frac{f(x)}{x-r}=\lim_{x\to r}\frac{f(x)-f(r)}{x-r}=f'(r),$$ by the definition of the derivative.

Answer 2

The claim you are trying to prove is wrong. For example, if $k=0$, the expression in the limit becomes $$\frac{x^n+ax+b}{x} = x^{n-1} + a + \frac bx$$ and the limit of this does not exist if $b\neq 0$.

In general, the limit will not exist whenever $k^n + ak + b$ is different from $0$, since in that case, the numerator tends to a nonzero number and the denominator tends to $0$.

If $k^n+ak+b=0$, then you can use that fact to write $x^n+ax+b$ as a product of $(x-k)$ and some other polynomial (remember, $k$ is a root of polynomial $p$ if and only if $p$ is divisible by $(x-k)$!)

Answer 3

Observe that we have

$$\lim_{x\to k}\frac{x^n+ax+b}{x-k}=nk^{n-1}+a \iff k^n+ak+b=0.$$

So , $\lim_{x\to k}\frac{x^n+ax+b}{x-k}=nk^{n-1}+a $ is not true , if $k^n+ak +b \ne 0.$