# How do you find the limit lim (3^x-2^x+1)/(4*3^x-2^x-1) as x->oo?

Jul 19, 2017

${\lim}_{x \rightarrow \infty} \frac{{3}^{x} - {2}^{x} + 1}{4 \cdot {3}^{x} - {2}^{x} - 1} = \frac{1}{4}$

#### Explanation:

We seek:

$L = {\lim}_{x \rightarrow \infty} \frac{{3}^{x} - {2}^{x} + 1}{4 \cdot {3}^{x} - {2}^{x} - 1}$

We can manipulate the limits as follows:

$L = {\lim}_{x \rightarrow \infty} \frac{{3}^{x} - {2}^{x} + 1}{4 \cdot {3}^{x} - {2}^{x} - 1} \cdot \frac{\frac{1}{3} ^ x}{\frac{1}{3} ^ x}$

$\setminus \setminus = {\lim}_{x \rightarrow \infty} \frac{\left(\frac{1}{3} ^ x\right) \left({3}^{x} - {2}^{x} + 1\right)}{\left(\frac{1}{3} ^ x\right) \left(4 \cdot {3}^{x} - {2}^{x} - 1\right)}$

$\setminus \setminus = {\lim}_{x \rightarrow \infty} \frac{{3}^{x} / {3}^{x} - {2}^{x} / {3}^{x} + \frac{1}{3} ^ x}{4 \cdot {3}^{x} / {3}^{x} - {2}^{x} / {3}^{x} - \frac{1}{3} ^ x}$

$\setminus \setminus = {\lim}_{x \rightarrow \infty} \frac{1 - {\left(\frac{2}{3}\right)}^{x} + {\left(\frac{1}{3}\right)}^{x}}{4 \cdot 1 - {\left(\frac{2}{3}\right)}^{x} - {\left(\frac{1}{3}\right)}^{x}}$

Now ${a}^{x} \rightarrow 0$ as $x \rightarrow \infty$ provided $| a | < 1$, And so:

$L = \frac{1 - 0 + 0}{4 - 0 - 0}$

$\setminus \setminus = \frac{1}{4}$