# How do you solve the limit as h approaches 0 of [34/(35+h) - (34/35)]/h?

Jun 26, 2015

${\lim}_{x \to 0} \frac{\frac{34}{35 + x} - \frac{34}{35}}{x} = - \frac{34}{1225} \approx - 0.028$.

#### Explanation:

${\lim}_{x \to 0} f \frac{x}{g} \left(x\right) = {\lim}_{x \to 0} \frac{\frac{34}{35 + x} - \frac{34}{35}}{x} = ' ' \left(\frac{0}{0}\right) ' '$, which is undefined.

We have $f \left(x\right) = \frac{34}{35 + x} - \frac{34}{35}$ and $g \left(x\right) = x$.

Since ${\lim}_{x \to 0} f \left(x\right) = {\lim}_{x \to 0} g \left(x\right) = 0$, we can use L'Hospital's Rule :

${\lim}_{x \to 0} f \frac{x}{g} \left(x\right) = {\lim}_{x \to 0} \frac{f ' \left(x\right)}{g ' \left(x\right)} = {\lim}_{x \to 0} \frac{\left[\frac{34}{35 + x} - \frac{34}{35}\right] '}{\left(x\right) '}$

$= {\lim}_{x \to 0} \frac{\left(\frac{34}{35 + x}\right) '}{\left(x\right) '}$

$= {\lim}_{x \to 0} \frac{\frac{- 34}{35 + x} ^ 2}{1} = {\lim}_{x \to 0} \frac{- 34}{35 + x} ^ 2 = - \frac{34}{35} ^ 2 = - \frac{34}{1225} \approx - 0.028$.

Jun 26, 2015

Rewrite (simplify) the expression.

#### Explanation:

Start by writing the numerator as a single ratio:

$\frac{34}{35 + h} - \frac{34}{35} = \frac{\left(34\right) \left(35\right) - \left(34\right) \left(35 + h\right)}{\left(35 + h\right) \left(35\right)}$

$= \frac{- 34 h}{35 \left(35 + h\right)}$

So the big ratio becomes:

$\frac{\frac{34}{35 + h} - \frac{34}{35}}{h} = \frac{\frac{\left(34\right) \left(35\right) - \left(34\right) \left(35 + h\right)}{\left(35 + h\right) \left(35\right)}}{\frac{h}{1}}$

$= \frac{- 34 h}{35 \left(35 + h\right)} \cdot \frac{1}{h}$

$= \frac{- 34}{35 \left(35 + h\right)}$

For the limit, we get:

${\lim}_{h \rightarrow 0} \frac{\frac{34}{35 + h} - \frac{34}{35}}{h} = {\lim}_{h \rightarrow 0} \frac{- 34}{35 \left(35 + h\right)} = \frac{- 34}{35} ^ 2$