Question

How do I find the limit of the expression lim h→0 √(1+h) - 1/h without using L'Hopital's Rule?

Answer 1

To evaluate the limit

\[
\lim_{h \to 0} \frac{\sqrt{1+h} - 1}{h}
\]

without using L'Hôpital's rule, we can use a technique called rationalizing the numerator. This involves multiplying by the conjugate of the numerator to simplify the expression.

1. **Rationalize the expression**:

Multiply the numerator and denominator by the conjugate of the numerator:

\[
\lim_{h \to 0} \frac{\sqrt{1+h} - 1}{h} \cdot \frac{\sqrt{1+h} + 1}{\sqrt{1+h} + 1}
\]

This gives us:

\[
= \lim_{h \to 0} \frac{(\sqrt{1+h} - 1)(\sqrt{1+h} + 1)}{h(\sqrt{1+h} + 1)}
\]

The numerator simplifies as follows:

\[
(\sqrt{1+h} - 1)(\sqrt{1+h} + 1) = (\sqrt{1+h})^2 - 1^2 = (1+h) - 1 = h
\]

Therefore, the limit becomes:

\[
= \lim_{h \to 0} \frac{h}{h(\sqrt{1+h} + 1)}
\]

2. **Simplify the expression**:

We can cancel \(h\) in the numerator and denominator (note: \(h \neq 0\) in the limit):

\[
= \lim_{h \to 0} \frac{1}{\sqrt{1+h} + 1}
\]

3. **Evaluate the limit**:

As \(h\) approaches 0, \(\sqrt{1+h}\) approaches \(\sqrt{1+0} = 1\):

\[
= \frac{1}{1 + 1} = \frac{1}{2}
\]

Thus, the limit is

\[
\boxed{\frac{1}{2}}.
\]