#### Answer

The statement is false.

#### Work Step by Step

$$\lim\limits_{x \to 1}\frac{x^2+6x-7}{x^2+5x-6}=\frac{\lim\limits_{x \to 1}(x^2+6x-7)}{\lim\limits_{x \to 1}(x^2+5x-6)}$$
Consider the denominator $$\lim\limits_{x \to 1}(x^2+5x-6)=1^2+5\times1-6=0$$
The quotient law states that $$\lim\limits_{x \to a}\frac{f(x)}{g(x)}=\frac{\lim\limits_{x \to a}f(x)}{\lim\limits_{x \to a}g(x)}$$ if $\lim\limits_{x \to a}g(x)\ne0$ (the limit of the denominator must not be 0)
However, in the above statement, we see that the limit of the denominator equals 0.
So, the statement is false.