Limit of the Reciprocal Function
The limit of a reciprocal function characterizes the behavior of $$ g(x) = \frac{1}{f(x)} $$ given the limiting behavior of $ f(x) $ in a neighborhood of a point \( x_0 \).
\[
\begin{array}{|c|c|}
\hline
\vphantom{\dfrac{1}{l}}
\textbf{Limit of } f(x) & \textbf{Limit of } \dfrac{1}{f(x)} \\
\hline
\vphantom{\dfrac{1}{l}}
l \in \mathbb{R},\; l \neq 0 & \dfrac{1}{l} \\
\hline
\vphantom{\dfrac{1}{l}}
+\infty & 0 \\
\hline
\vphantom{\dfrac{1}{l}}
-\infty & 0 \\
\hline
\vphantom{\dfrac{1}{l}}
0^{+} & +\infty \\
\hline
\vphantom{\dfrac{1}{l}}
0^{-} & -\infty \\
\hline
\end{array}
\]
The key idea is immediate. The limit of the reciprocal function depends on the value approached by $ f(x) $, with special attention to whether that value is zero and to its sign.
Three principal scenarios arise.
Finite Limit Different from Zero
If $ f(x) $ admits a finite, nonzero limit $$ \lim_{x \to x_0} f(x) = l \in \mathbb{R}, \quad l \neq 0 $$ then the reciprocal converges to the reciprocal of that limit $$ \lim_{x \to x_0} \frac{1}{f(x)} = \frac{1}{l} $$
Equivalently, as $ f(x) $ approaches a real number distinct from zero, \( 1/f(x) \) approaches \( 1/l \).
Example
Evaluate
$$ \lim_{x \to 1} \frac{1}{2x + 1} $$
The denominator is
$$ f(x) = 2x + 1 $$
First compute
$$ \lim_{x \to 1} (2x+1) = 3 $$
Because the limit equals 3, which is nonzero, the theorem applies directly.
$$ \lim_{x \to 1} \frac{1}{2x+1} = \frac{1}{3} $$
The conclusion follows at once.
Infinite Limit
If $ f(x) $ diverges $$ \lim_{x \to x_0} f(x) = +\infty \quad \text{or} \quad -\infty $$ and there exists a neighborhood of \( x_0 \) in which \( f(x) \neq 0 \), then the reciprocal tends to zero $$
\lim_{x \to x_0} \frac{1}{f(x)} = 0 $$
This reflects a basic numerical fact. The reciprocal of a quantity whose magnitude increases without bound becomes arbitrarily small.
$$ \frac{1}{1000} = 0.001 $$
$$ \frac{1}{1,000,000} = 0.000001 $$
$$ \vdots $$
As the denominator grows, the reciprocal approaches zero.
Example
Consider
$$ \lim_{x \to \infty} \frac{1}{x^2} $$
Here
$$ f(x) = x^2 $$
Then
$$ \lim_{x \to \infty} x^2 = + \infty $$
Hence
$$ \lim_{x \to \infty} \frac{1}{x^2} = 0 $$
The result is consistent with the theorem.
Limit Approaching Zero
If $ f(x) $ approaches zero $$ \lim_{x \to x_0} f(x) = 0^+ \quad \text{or} \quad 0^- $$ the reciprocal diverges, with the sign determined by the direction of approach.
- If $$ \lim_{x \to x_0} f(x) = 0^+ $$ then $$ \lim_{x \to x_0} \frac{1}{f(x)} = +\infty $$
- If $$ \lim_{x \to x_0} f(x) = 0^- $$ then $$ \lim_{x \to x_0} \frac{1}{f(x)} = -\infty $$
When the denominator becomes arbitrarily small in magnitude, the reciprocal becomes arbitrarily large in magnitude. The sign is decisive because the reciprocal preserves it.
If the limit approaches positive zero \( 0^+ \), the reciprocal diverges to plus infinity.
$$ \frac{1}{0.1} = 10 $$
$$ \frac{1}{0.001} = 1000 $$
$$ \vdots $$
If the limit approaches negative zero \( 0^- \), the reciprocal diverges to minus infinity.
$$ \frac{1}{-0.1} = -10 $$
$$ \frac{1}{-0.001} = -1000 $$
$$ \vdots $$
Example
Evaluate
$$ \lim_{x \to 0^+} \frac{1}{x^3} $$
The denominator is
$$ f(x) = x^3 $$
Compute
$$ \lim_{x \to 0^+} x^3 = 0^+ $$
The function approaches positive zero because values of \( x \) near zero from the right are positive.
Therefore
$$ \lim_{x \to 0^+} \frac{1}{x^3} = + \infty $$
Example 2
Evaluate
$$ \lim_{x \to 1^+} \frac{1}{1-x} $$
The denominator is
$$ f(x) = 1-x $$
Compute
$$ \lim_{x \to 1^+} 1-x = 0^- $$
The function approaches negative zero because values slightly greater than 1 make \( 1-x \) negative.
Hence
$$ \lim_{x \to 1^+} \frac{1}{1-x} = - \infty $$
And so forth.
