Question

Find the limit of the function algebraically. (2 points)

Answer 1

$\lim\limits_{x\to9}\dfrac{x^2-81}{x-9}=\lim\limits_{x\to9}\dfrac{x^2-9^2}{x-9}=\lim\limits_{x\to9}\dfrac{(x+9)(x-9)}{x-9}=\lim\limits_{x\to9}(x+9)=\\\\\\=9+9=\boxed{18}$

Answer 2

Answer:  The required value of the limit is 18.

Step-by-step explanation:  We are given to find the limit of the following function algebraically :

limit as x approaches nine of quantity x squared minus eighty one divided by quantity x minus nine.

We will be using the following factorization formula :

$a^2-b^2=(a+b)(a-b).$

The limit can be calculated as follows :

$\ell\\\\\\=\lim_{x\rightarrow 81}\dfrac{x^2-81}{x-9}\\\\\\=\lim_{x\rightarrow 9}\dfrac{x^2-9^2}{x-9}\\\\\\=\lim_{x\rightarrow 9}\dfrac{(x+9)(x-9)}{(x-9)}\\\\=\lim_{x\rightarrow 9}(x+9)~~~~~~~~~~~~~~~~[\textup{since }x\rightarrow 9,~so~x\neq 9]\\\\=9+9\\\\=18.$

Thus, the required value of the limit is 18.