Question

find the limit.
\\( \lim\limits_{h \to 0} \frac{\sqrt{18h + 1} - 1}{h} \\)

select the correct choice below and, if necessary, fill in the answer box to complete your choice.
\\( \bigcirc \\) a. \\( \lim\limits_{h \to 0} \frac{\sqrt{18h + 1} - 1}{h} = \square \\) (type an integer or a simplified fraction.)
\\( \bigcirc \\) b. the limit does not exist.

Explanation:

Step1: Rationalize the numerator

Multiply the numerator and denominator by the conjugate of the numerator, which is $\sqrt{18h + 1}+1$.

$$\lim_{h ightarrow0}\frac{\sqrt{18h + 1}-1}{h}\times\frac{\sqrt{18h + 1}+1}{\sqrt{18h + 1}+1}$$

Using the difference of squares formula $(a - b)(a + b)=a^{2}-b^{2}$ in the numerator, we get:

$$\lim_{h ightarrow0}\frac{(\sqrt{18h + 1})^{2}-1^{2}}{h(\sqrt{18h + 1}+1)}$$

Simplify the numerator:

$$\lim_{h ightarrow0}\frac{18h + 1 - 1}{h(\sqrt{18h + 1}+1)}$$

Step2: Simplify the expression

Simplify the numerator by canceling out $1$ and $-1$:

$$\lim_{h ightarrow0}\frac{18h}{h(\sqrt{18h + 1}+1)}$$

Cancel out the common factor $h$ in the numerator and the denominator (since $h
ightarrow0$ but $h
eq0$ at the limit process):

$$\lim_{h ightarrow0}\frac{18}{\sqrt{18h + 1}+1}$$

Step3: Evaluate the limit

Now we can substitute $h = 0$ into the simplified expression:
$$\frac{18}{\sqrt{18\times0 + 1}+1}=\frac{18}{1 + 1}=\frac{18}{2}=9$$

Answer:

A. $\lim\limits_{h
ightarrow0}\frac{\sqrt{18h + 1}-1}{h}=\boxed{9}$