Find $a$ such that $\lim \limits_{x \to 1 } {{\frac{x^2+(3-a)x+3a}{(x-1)}}}= 7$

Question

I have this limit equation below:

$$ \lim \limits_{x \to 1 } {{\frac{x^2+(3-a)x+3a}{(x-1)}}}= 7 $$

One needs to determine the value of $a$ (where $a$ is a real number) such that as $x$ approaches $1$, the limit becomes the value $7$.

My question is how to approach this. I wondering if there is a algebraic computation approach to this, or is it more of trying to figure out values of $a$ such that the top factors in factors that would cancel out the bottom denominator and we get a linear term such that we get the value of $7$ in the end.

Just curious to know if there is more than one way to do this problem.

Answer 1

Since the limit is finite, and the denominator tends to zero, the numerator must also tend to zero as $x \to 1$. This means that $$ 1^2+(3-a) \cdot 1 + 3a = 0 \Leftrightarrow a= - 2 $$

This means that this can only work if $a=-2$. Now we must check that it actually works...

$$ \lim_{x \to 1} \dfrac{x^2+5x-6}{x-1} = \lim_{x\to 1} \dfrac{(x-1)(x+6)}{x-1} = \lim_{x\to 1} (x+6)=7. $$

Answer 2

Hint: Use that $$\frac{x^2+(3-a)x+3a}{x-1}=-a+x+4+\frac{2(a+2)}{x-1}$$

Answer 3

The following limit becomes infinite due to the denominator. But the RHS says that it is a finite number. It can only happen if the numerator polynomial is having $(x-1)$ as a factor so that the factor would cancel out with the denominator to give a finite number.

So, the numerator is having $(x-1)$ as a factor, using factor theorem on $f(x)=x^2+(3-a)x+3a$, $$f(1)=0$$ $$1+(3-a)(1)+3a=0$$ $$2a=-4$$ $$a=-2$$ You can check plugging $a=-2$ in the numerator and solving the limit.

Hope this helps...