# How do you find the limit of ((h-1)^2+1)/h as h approaches 0?

May 20, 2017

${\lim}_{h \to 0} \frac{{\left(h - 1\right)}^{2} + 1}{h} = \infty$

#### Explanation:

Plugging in $0$ for $h$ gives that:

${\lim}_{h \to 0} \frac{{\left(h - 1\right)}^{2} + 1}{h} = \frac{{\left(0 - 1\right)}^{2} + 1}{0} = \frac{2}{0}$

Since $\frac{2}{0}$ equates to the indeterminate form $\infty$, we can say that ${\lim}_{h \to 0} \frac{{\left(h - 1\right)}^{2} + 1}{h} = \infty$.

May 20, 2017

The limit is divergent.

#### Explanation:

The whole point of evaluating a limit is that we do not generally look (or care) at the behaviour when $h = 0$, just when $h \rightarrow 0$

In this case we have:

${\lim}_{h \rightarrow 0} \frac{{\left(h - 1\right)}^{2} + 1}{h} = {\lim}_{h \rightarrow 0} \frac{\left({h}^{2} - 2 h + 1\right) + 1}{h}$
$\text{ } = {\lim}_{h \rightarrow 0} \frac{{h}^{2} - 2 h + 2}{h}$
$\text{ } = {\lim}_{h \rightarrow 0} \left(h - 2 + \frac{2}{h}\right)$
$\text{ } = {\lim}_{h \rightarrow 0} h - {\lim}_{h \rightarrow 0} 2 + 2 {\lim}_{h \rightarrow 0} \frac{1}{h}$

The First limit exists:

${\lim}_{h \rightarrow 0} h = 0$

The Second limit exists

${\lim}_{h \rightarrow 0} 2 = 2$

However the third limit diverges

$2 {\lim}_{h \rightarrow 0} \frac{1}{h} \rightarrow \infty$

As $\frac{1}{h}$increases without bound as $h \rightarrow 0$

Hence the initial limit is divergent.