Question

find the limit of:
(2^(2x)+2^x-20))/(2^x-4)

Answer 1

find the limit of that

Answer 2

As \(x\) approaches...

I'll assume you mean \(x\to2\).
\[\lim_{x\to2}\frac{2^{2x}+2^x-20}{2^x-4}\]
Let's see what happens if we were to substitute \(t=2^x\). Then, as \(x\to2\), you have \(t\to4\), and the limit can be rewritten as
\[\lim_{t\to4}\frac{t^2+t-20}{t-4}\]
Can you solve this one?

Answer 3

yes now i can but what rule did you use where you just chaged when x goes to 2 to 4?

Answer 4

When you make the substitution, you have to consider what happens to \(t\) as \(x\to2\). Since I define \(t=2^x\), as \(x\to2\), you have \(t\to2^2\), or \(t\to4\).

Note that the substitution isn't necessary, but it somewhat helps you notice that you basically have a quadratic expression in the numerator that you can factor.

If you want to keep on using \(x\), you do just that:
\[\lim_{x\to2}\frac{2^{2x}+2^x-20}{2^x-4}=\lim_{x\to2}\frac{\left(2^x+5\right)\left(2^x-4\right)}{2^x-4}=\lim_{x\to2}\left(2^x-4\right)\]

Answer 5

oh okay thanks that makes seance now.