Limit of (e^x − e^(tan x))/(x − tan x)
Evaluate the limit: $$ L = \lim_{x \to 0} \frac{e^x - e^{\tan x}}{x - \tan x}. $$
Explanation of Method
As $x \to 0$, both $x \to 0$ and $\tan x \to 0$, so the numerator $e^x - e^{\tan x} \to e^0 - e^0 = 0$ and the denominator $x - \tan x \to 0$, giving the indeterminate form $\dfrac{0}{0}$. Rather than differentiating repeatedly, it is cleaner to factor the numerator around the common exponential term and reduce the problem to the basic limit $\lim_{u\to 0}\dfrac{e^u-1}{u} = 1$.
- Introduce the small quantity $d = x - \tan x$, which tends to $0$.
- Factor the numerator as $e^{\tan x}\left(e^{\,x-\tan x} - 1\right)$.
- Apply the standard limit $\lim_{u\to 0}\dfrac{e^u - 1}{u} = 1$.
- Combine with the limit of the remaining exponential factor.
Step 1 · Introduce the difference $d = x - \tan x$
Let $d = x - \tan x$. As $x \to 0$, both $x \to 0$ and $\tan x \to 0$, so $d \to 0$ as well. The exponent in the numerator can then be written as $x = \tan x + d$, so:
$$ e^x = e^{\tan x + d} = e^{\tan x} \cdot e^{d}. $$
Step 2 · Factor the numerator
Using this, the numerator becomes:
$$ e^x - e^{\tan x} = e^{\tan x} \cdot e^{d} - e^{\tan x} = e^{\tan x}\left(e^{d} - 1\right). $$
Since the denominator is $x - \tan x = d$, the whole expression is:
$$ \frac{e^x - e^{\tan x}}{x - \tan x} = \frac{e^{\tan x}\left(e^{d}-1\right)}{d} = e^{\tan x} \cdot \frac{e^{d}-1}{d}. $$
Step 3 · Evaluate each factor as $x \to 0$
For the first factor, since $\tan x \to 0$ as $x \to 0$:
$$ \lim_{x \to 0} e^{\tan x} = e^0 = 1. $$
For the second factor, recall the standard limit $\displaystyle\lim_{u \to 0} \frac{e^{u}-1}{u} = 1$ (the derivative of $e^u$ at $u=0$). Since $d = x-\tan x \to 0$ as $x \to 0$:
$$ \lim_{x \to 0} \frac{e^{d}-1}{d} = 1. $$
Step 4 · Combine the results
Multiplying the two limits:
$$ L = \lim_{x \to 0} e^{\tan x} \cdot \lim_{x \to 0}\frac{e^{d}-1}{d} = 1 \cdot 1 = 1. $$
Summary Table
| Factor | Rewritten as | Limit as $x \to 0$ |
|---|---|---|
| $e^{\tan x}$ | exponential of $\tan x \to 0$ | $1$ |
| $\dfrac{e^{d}-1}{d}$, $d = x-\tan x$ | standard limit at $d=0$ | $1$ |
Additional Notes
- This factoring trick avoids differentiating $e^{\tan x}\sec^2 x$ repeatedly, which quickly becomes messy if L'Hôpital's Rule is applied more than once.
- The key identity used is $e^a - e^b = e^b\left(e^{a-b}-1\right)$, which isolates the difference $a-b$ so the standard limit $\lim_{u\to 0}\frac{e^u-1}{u}=1$ can be applied directly.
- The same method works for any limit of the form $\lim \dfrac{e^{g(x)}-e^{h(x)}}{g(x)-h(x)}$ where $g(x), h(x) \to c$ for the same value $c$: the limit is always $e^{c}$.
Result
The value of the limit is: