Find the limit of (2^x - 3^x)/(2^x + 3^x) as x approaches infinity.

Step-by-step solution

1. 1

   Diagnose the form. Both $2^x$ and $3^x$ tend to $\infty$, so the quotient is $\tfrac{\infty-\infty}{\infty+\infty}$ — indeterminate. Note that l’Hôpital would be a poor choice here: differentiating reproduces the same shape with extra $\ln 2$ and $\ln 3$ factors and never simplifies.

2. 2

   Decide which exponential dominates. For large $x$, comparing bases decides everything: since $3>2$,

   $\frac{2^{x}}{3^{x}}=\left(\frac{2}{3}\right)^{x}\longrightarrow 0\quad\text{as }x\to\infty$

   because the base $\tfrac23<1$. So $3^x$ is the dominant term in both numerator and denominator.

3. 3

   Divide numerator and denominator by the dominant term $3^{x}$. This is the key algebraic step — dividing by the largest growing quantity is what removes the indeterminacy:

   $\frac{2^{x}-3^{x}}{2^{x}+3^{x}}=\frac{\left(\frac23\right)^{x}-1}{\left(\frac23\right)^{x}+1}$

4. 4

   Take the limit. Writing $t=\left(\tfrac23\right)^{x}\to 0^{+}$:

   $\lim_{t\to 0^{+}}\frac{t-1}{t+1}=\frac{0-1}{0+1}=-1$

   So the limit is $-1$. Intuitively the $-3^x$ swamps everything, and the ratio approaches $\tfrac{-3^x}{3^x}=-1$.

5. 5

   Check the approach numerically and consider the other direction. At $x=50$ the quotient is $-0.9999999969$ and by $x=200$ it is $-1$ to machine precision $\checkmark$. The approach is from above, since $\tfrac{t-1}{t+1}<0$ but increases toward $-1$ as $t\to 0$. For contrast, as $x\to-\infty$ the roles reverse — $2^x$ dominates — and the same expression tends to $+1$.