AP Calculus AB and BC

Limit of 3^x/4^x at Infinity Is 0

Q: What is the limit of 4^x/3^x?

It is \infty. The same collapse gives \left(\frac{4}{3}\right)^{x}, and a base greater than 1 grows without bound. Flipping the fraction flips the answer, which is what you expect when the original limit is 0 and the function stays positive.

Q: Do I need L'Hopital's rule for this?

No, and it does not help. Applying it once gives \frac{3^x \ln 3}{4^x \ln 4}, which is the original quotient multiplied by the constant \frac{\ln 3}{\ln 4}, so the form never resolves. Combining the bases first is the intended route.

Q: What happens as x approaches negative infinity?

The limit is \infty. Negative exponents invert the base, so \left(\frac{3}{4}\right)^{x} becomes \left(\frac{4}{3}\right)^{|x|}, which grows without bound. At x = -10 the value is already about 17.76.

The limit of 3^x over 4^x as x approaches infinity is 0. Substitution gives infinity over infinity, but no calculus is needed: a shared exponent means the quotient is (3/4)^x, and a base strictly between 0 and 1 raised to a growing exponent decays to 0. Comparing the two bases is the whole method.

\lim_{x \to \infty} \frac{3^x}{4^x} = 0

Settled by combining into a single base.

One base instead of two

The two powers share the same exponent, so the exponent laws collect them into a single power and the indeterminate form disappears before any calculus starts.

\frac{3^x}{4^x} = \left(\frac{3}{4}\right)^{x}

The base $\frac{3}{4}$ is less than $1$ , so a growing exponent means multiplying by $0.75$ over and over, which drives the value toward $0$ .

\lim_{x \to \infty} \left(\frac{3}{4}\right)^{x} = 0 \qquad \left(0 < \frac{3}{4} < 1\right)

Rewriting the base as an exponential makes the reason explicit. Since $\ln\frac{3}{4} \approx -0.2877$ is negative, the exponent is heading to $-\infty$ .

\left(\frac{3}{4}\right)^{x} = e^{x\ln\frac{3}{4}} \longrightarrow 0

$x$	$3^x$	$4^x$	$\frac{3^x}{4^x}$
$10$	$59{,}049$	$1{,}048{,}576$	$0.05631$
$20$	$3.487 \cdot 10^{9}$	$1.100 \cdot 10^{12}$	$0.003171$
$50$	$7.179 \cdot 10^{23}$	$1.268 \cdot 10^{30}$	$0.00000057$

Both columns explode, and the ratio still collapses. That is the point of the form being indeterminate.

Why substitution fails

Both $3^x$ and $4^x$ are unbounded, so substituting gives $\frac{\infty}{\infty}$ . The form says the two are large; the answer depends on which is larger and by what factor, and the factor is $\left(\frac{3}{4}\right)^{x}$ , which is not a constant.

L'Hopital's rule applies and gets nowhere. Each exponential differentiates to itself times the log of its base, so one pass returns the same quotient scaled by $\frac{\ln 3}{\ln 4} \approx 0.79$ , and every later pass multiplies by that constant again.

\lim_{x \to \infty} \frac{3^x}{4^x} \;\longrightarrow\; \lim_{x \to \infty} \frac{3^x\ln 3}{4^x \ln 4} = \frac{\ln 3}{\ln 4}\lim_{x \to \infty} \frac{3^x}{4^x}

Exponentials never compromise

A ratio of two exponentials with different bases is always $0$ or $\infty$ , never a finite nonzero number. The larger base eventually beats the smaller one by any margin you care to name, so the only question is which base is larger. That is why the comparison replaces the calculus here.

The rule for r to the x

\lim_{x \to \infty} r^{x} = \begin{cases} \infty & r > 1 \\ 1 & r = 1 \\ 0 & 0 < r < 1 \end{cases}

With $r = \frac{3}{4}$ the third line applies. Nothing about the original bases matters beyond their ratio, so $\frac{0.3^x}{0.4^x}$ has the same limit $0$ , and $\frac{300^x}{400^x}$ does too.

The other direction reverses the roles, since $\left(\frac{3}{4}\right)^{x} = \left(\frac{4}{3}\right)^{-x}$ and the exponent $-x$ grows.

\lim_{x \to -\infty} \frac{3^x}{4^x} = \infty

The mistakes students make

Cancelling the exponents to get $\frac{3}{4}$ . The $x$ is not a factor to cross out, and $\left(\frac{3}{4}\right)^{x}$ equals $\frac{3}{4}$ only at $x = 1$ .
Answering $\infty$ because both parts are unbounded. Both are, and the smaller base loses anyway.
Reading it as $\frac{x^3}{x^4}$ . There the base varies and the exponent is fixed, which is the opposite arrangement, and the reasoning that settles one does not settle the other.
Deciding that $\left(\frac{3}{4}\right)^{x}$ levels off at some small positive number. Each unit step multiplies by $0.75$ again, so there is no floor to level off at.
Running L'Hopital's rule repeatedly. It is legal and it reproduces the original quotient times $\frac{\ln 3}{\ln 4}$ every time.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

What is the limit of 4^x/3^x?

It is $\infty$ . The same collapse gives $\left(\frac{4}{3}\right)^{x}$ , and a base greater than $1$ grows without bound. Flipping the fraction flips the answer, which is what you expect when the original limit is $0$ and the function stays positive.

Do I need L'Hopital's rule for this?

No, and it does not help. Applying it once gives $\frac{3^x \ln 3}{4^x \ln 4}$ , which is the original quotient multiplied by the constant $\frac{\ln 3}{\ln 4}$ , so the form never resolves. Combining the bases first is the intended route.

What happens as x approaches negative infinity?

The limit is $\infty$ . Negative exponents invert the base, so $\left(\frac{3}{4}\right)^{x}$ becomes $\left(\frac{4}{3}\right)^{|x|}$ , which grows without bound. At $x = -10$ the value is already about $17.76$ .