AP Calculus AB and BC

Limit of (e^(2x)-1)/x as x Approaches 0 Is 2

Q: Why does the 2 survive when the exponential itself goes to 1?

Because the numerator and the denominator both collapse to 0, so the limit compares speeds, not values. Near 0 the numerator behaves like 2x, since e^{2x} \approx 1 + 2x, so it shrinks half as fast as it would with exponent x. Doubling the exponent doubles the slope, and the limit records the slope.

Q: What is \lim_{x \to 0} \frac{e^{kx}-1}{x} in general?

It is k, for any constant, positive or negative. The same multiply-and-divide step gives k \cdot \frac{e^{kx}-1}{kx}, whose second factor tends to 1. So \frac{e^{-3x}-1}{x} \to -3. If the denominator is mx rather than x, the answer is \frac{k}{m}.

Q: Does the answer change if the denominator is 2x?

Yes, it becomes 1. With denominator 2x the exponent and the denominator already match, so \lim_{x \to 0}\frac{e^{2x}-1}{2x} = 1 and no adjustment is needed. Checking whether they match is the whole of the technique.

The limit of (e^(2x) - 1)/x as x approaches 0 is 2, not 1. Direct substitution gives 0/0. Multiply and divide by 2 to rebuild the standard form, 2 times (e^(2x) - 1)/(2x), whose second factor tends to 1. It is also the derivative of e^(2x) at 0, which is 2 by the chain rule.

\lim_{x \to 0} \frac{e^{2x}-1}{x} = 2

Settled by the limit definition of the derivative.

Rebuilding the standard form

The known result is $\lim_{u \to 0}\frac{e^u-1}{u} = 1$ , where the exponent and the denominator are the same thing. Here the exponent is $2x$ while the denominator is only $x$ , so repair the mismatch by multiplying and dividing by $2$ .

\frac{e^{2x}-1}{x} = \frac{2}{2}\cdot\frac{e^{2x}-1}{x} = 2 \cdot \frac{e^{2x}-1}{2x}

Now set $u = 2x$ . As $x \to 0$ the new variable also goes to $0$ , so the fraction on the right is the standard limit and contributes $1$ .

\lim_{x \to 0} \frac{e^{2x}-1}{x} = 2\lim_{u \to 0} \frac{e^{u}-1}{u} = 2 \cdot 1 = 2

A factoring route to the same 2

Since $e^{2x} = \left(e^x\right)^2$ , the numerator is a difference of squares, $\left(e^x-1\right)\left(e^x+1\right)$ . Dividing by $x$ splits it as $\frac{e^x-1}{x}\cdot\left(e^x+1\right)$ , and the two factors tend to $1$ and $2$ .

The same limit as a derivative

Set $f(t) = e^{2t}$ , so $f(0) = 1$ and the fraction is the difference quotient for $f$ based at $0$ . The limit is $f'(0)$ , and the chain rule is what supplies the inner coefficient.

f'(t) = 2e^{2t}, \qquad f'(0) = 2e^{0} = 2

That is the structural reason the answer is not $1$ . The graph of $y = e^{2x}$ crosses the $y$ -axis at the same height as $y = e^x$ but twice as steeply, and this limit reads off the steepness.

What direct substitution gives

At $x = 0$ the exponent vanishes, so the numerator is $1 - 1 = 0$ while the denominator is $0$ .

\frac{e^{2 \cdot 0}-1}{0} = \frac{1-1}{0} = \frac{0}{0}

The form is indeterminate, so the value is still open. L'Hopital's rule is available and lands on the same number in one pass, because differentiating the top brings the $2$ down.

\lim_{x \to 0} \frac{e^{2x}-1}{x} = \lim_{x \to 0} \frac{2e^{2x}}{1} = 2

The mistakes students make

Answering $1$ from memory of $\frac{e^x-1}{x}$ . The exponent and the denominator have to match before that result can be quoted.
Answering $\frac{1}{2}$ by dividing where you should multiply. Compensating for a denominator of $2x$ needs a factor of $2$ out front.
Writing $e^{2x} = 2e^x$ . The correct rearrangement is $e^{2x} = \left(e^x\right)^2$ .
Cancelling the $x$ in the exponent against the $x$ underneath. An exponent is not a factor, so nothing cancels.

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

Why does the $2$ survive when the exponential itself goes to $1$ ?

Because the numerator and the denominator both collapse to $0$ , so the limit compares speeds, not values. Near $0$ the numerator behaves like $2x$ , since $e^{2x} \approx 1 + 2x$ , so it shrinks half as fast as it would with exponent $x$ . Doubling the exponent doubles the slope, and the limit records the slope.

What is $\lim_{x \to 0} \frac{e^{kx}-1}{x}$ in general?

It is $k$ , for any constant, positive or negative. The same multiply-and-divide step gives $k \cdot \frac{e^{kx}-1}{kx}$ , whose second factor tends to $1$ . So $\frac{e^{-3x}-1}{x} \to -3$ . If the denominator is $mx$ rather than $x$ , the answer is $\frac{k}{m}$ .

Does the answer change if the denominator is $2x$ ?

Yes, it becomes $1$ . With denominator $2x$ the exponent and the denominator already match, so $\lim_{x \to 0}\frac{e^{2x}-1}{2x} = 1$ and no adjustment is needed. Checking whether they match is the whole of the technique.