Question

Does this limit exist? Show steps

Answer 1

Answer:

Yes, the limit does exist.

Step-by-step explanation:

By definition, a limit in the form:

$\displaystyle \lim_{x\to a}f(x)$

Exists if and only if:

$\displaystyle \lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)$

We are given the limit:

$\displaystyle \lim_{x\to 2}\frac{x^2+x-6}{x^2-4}$

So, in order to prove that this limit exists, we simply have to show that:

$\displaystyle \lim_{x\to 2^-}\frac{x^2+x-6}{x^2-4}=\displaystyle \lim_{x\to 2^+}\frac{x^2+x-6}{x^2-4}$

Let's do the left-hand side first. We can factor the function:

$=\displaystyle \lim_{x\to 2^-}\frac{(x+3)(x-2)}{(x+2)(x-2)}$

Simplify:

$\displaystyle =\lim_{x\to 2^-}\frac{x+3}{x+2}$

By direct substitution:

$\displaystyle =\frac{(2)+3}{(2)+2}=\frac{5}{4}$

Now, we can evaluate the right-hand side. Again, factor:

$=\displaystyle \lim_{x\to 2^+}\frac{(x+3)(x-2)}{(x+2)(x-2)}$

Simplify:

$\displaystyle =\lim_{x\to 2^+}\frac{x+3}{x+2}$

And by direct substitution:

$\displaystyle =\frac{(2)+3}{(2)+2}=\frac{5}{4}$

Since both the left- and right-hand limits equal 5/4, our original limit does indeed exist.