Question

$\left(a\right).$What is wrong with the following equation?

$\frac{x^2+x-6}{x-2}=x+3$

$\left(b\right).$In view of part$\left(a\right)$, explain why the equation

$\underset{x\to 2}{\lim }\frac{x^2+x-6}{x-2}=\underset{x\to 2}{\lim }(x+3)$

is correct.

Short answer

The equation $\frac{x^2+x-6}{x-2}=x+3$ is incorrect.

While the equation$\underset{x\to 2}{\lim }\frac{x^2+x-6}{x-2}=\underset{x\to 2}{\lim }(x+3)$ is correct.

Step-by-step solution

Step 1. Given information.

The equations here give are,

$\frac{x^2+x-6}{x-2}=x+3$$and$$\underset{x\to 2}{\lim }\frac{x^2+x-6}{x-2}=\underset{x\to 2}{\lim }(x+3)$

$\left(a\right)$

Step 2. Simplification.

Given, $\frac{x^2+x-6}{x-2}=x+3$

Then, apply factorization method

We get,

$\begin{array}{c}\frac{(x+3)+(x-2)}{(x-2)}=(x+3)\\ \frac{x+3}{1}=x+3\\ x+3=x+3\\ 0=0\end{array}$

Doesn’t give any sense so it is incorrect.

$\left(b\right)$

Step 3. Find the limits.

Given, $\underset{x\to 2}{\lim }\frac{x^2+x-6}{x-2}=\underset{x\to 2}{\lim }(x+3)$

Solve for L.H.S.,

$\begin{array}{c}L.H.S.=\underset{x\to 2}{\lim }\frac{x^2+x-6}{x-2}\\ =\underset{x\to 2}{\lim }\frac{(x+3)(x-2)}{(x-2)}\end{array}$

Cancellation of the common term $x-2$,

$\begin{array}{c}=\underset{x\to 2}{\lim }(x+3)\\ =5\end{array}$

Solve for R.H.S.,

$R.H.S.=\underset{x\to 2}{\lim }(x+3)$

Put the value $x=2$,

$\begin{array}{c}=2+3\\ =5\end{array}$

Here we see that, L.H.S.=R.H.S.

So, it is correct.