Lynch Long
05/26/2024 · Senior High School
\( \lim _ { h \rightarrow 0 } [ \frac { 1 } { h } \left( \begin{array} { c } { 1 } \\ { x + h } \end{array} \right) ] \)
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To evaluate the limit
\[
\lim_{h \rightarrow 0} \left[ \frac{1}{h} \binom{1}{x+h} \right],
\]
we first need to understand the binomial coefficient \(\binom{1}{x+h}\). The binomial coefficient \(\binom{n}{k}\) is defined as:
\[
\binom{n}{k} = \frac{n!}{k!(n-k)!}.
\]
For \(n = 1\), we have:
\[
\binom{1}{k} = \frac{1!}{k!(1-k)!} = \frac{1}{k!(1-k)!}.
\]
However, this definition is only valid for non-negative integers \(k\). For non-integer values, we can use the Gamma function, where \(\binom{n}{k} = \frac{\Gamma(n+1)}{\Gamma(k+1)\Gamma(n-k+1)}\).
Thus, we can express \(\binom{1}{x+h}\) as:
\[
\binom{1}{x+h} = \frac{1}{(x+h)! (1 - (x+h))!}.
\]
However, this expression becomes complicated for non-integer values. Instead, we can use the property of binomial coefficients in terms of the Beta function:
\[
\binom{1}{x+h} = \frac{1}{(x+h)(1 - (x+h))} = \frac{1}{(x+h)(1-x-h)}.
\]
Now, we can rewrite the limit:
\[
\lim_{h \rightarrow 0} \left[ \frac{1}{h} \binom{1}{x+h} \right] = \lim_{h \rightarrow 0} \left[ \frac{1}{h} \cdot \frac{1}{(x+h)(1-x-h)} \right].
\]
As \(h\) approaches \(0\), we can substitute \(h = 0\):
\[
\frac{1}{(x+0)(1-x-0)} = \frac{1}{x(1-x)}.
\]
Thus, we have:
\[
\lim_{h \rightarrow 0} \left[ \frac{1}{h} \cdot \frac{1}{(x+h)(1-x-h)} \right] = \lim_{h \rightarrow 0} \frac{1}{h} \cdot \frac{1}{x(1-x)}.
\]
This limit diverges unless \(x(1-x) = 0\), which occurs when \(x = 0\) or \(x = 1\).
Therefore, the limit is:
\[
\begin{cases}
\infty & \text{if } x \neq 0, 1 \\
\text{undefined} & \text{if } x = 0 \text{ or } x = 1.
\end{cases}
\]
In conclusion, the limit diverges for most values of \(x\) except at the endpoints \(x = 0\) and \(x = 1\).
Quick Answer
The limit diverges unless \(x = 0\) or \(x = 1\).
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