Solving Limits with Elimination

With solving limits via elimination, we are tipically analyzing a rational function that needs factoring of a term out of the polynomials on the top and/or the bottom to get out of the indeterminate form \((\frac{0}{0})\).

• Try factoring both the top and bottom
  • \((x\pm1)\)
  • \((x\pm2)\)
• Rationalize all of the square roots

Tip for picking factors

---

Tip! If you plug in a value to an expression, and out pops 0, that value is a zero of the expression. It is helpful like this

Factor: \((x^6-1)\)

As you could see, plugging \(x=1\) yields \(0\), meaning that \((x-1)\) is a zero of \((x^6-1)\), and hence would be able to be factored out.

To factor it out, either do synthetic division or long division.

---

Let's do a problem solve for \(\lim_{x\to 2} \frac{(x^2-4)}{(x-2)}\)

1. First, notice the fact this function will have a hole at \(x=2\). This is especially important because after we simplify we will loose this hole.
2. Ok, now let's simply. \(\frac{(x^2-4)}{(x-2)} = \frac{(x+2)(\cancel{(x-2)})}{(\cancel{x-2})} = (x+2)\)
3. Great! So, we know that this function behaves linearly with simply a hole at 2.
4. Doing the double-sided limits…
   • Evaluating \(\lim_{x\to2^+}\), the value will be \(4\) because \(2+2=4\).
   • Evaluating
     

Here's another one! \(\lim_{x\to0} \frac{\sqrt{x+4}-2}{x}\)

1. First, notice that if we are going to solve this problem, we have to divide the top thing (\(\sqrt{x+4}-2\)) by \(x\), somehow
2. The only thing we could do here is rationalize the top by multiplying the whole faction by a fancy one \(\frac{\sqrt{x+4}+2}{\sqrt{x+4}+2}\).
3. This results in \(\frac{x+4-4}{x\times(\sqrt{x+4}+2)}\), which simplifies to \(\frac{\cancel{x}}{\cancel{x}\times(\sqrt{x+4}+2)}\)
4. Plugging in \(x=0\), you get \(\frac{1}{4}\).

If there is no factors, you got yourself a vertical asymtote. Refer to #missing #disorganized for solution!