# How do you find lim ln(t^2+1)/t as t->0 using l'Hospital's Rule?

Dec 11, 2016

${\lim}_{t \rightarrow 0} \ln \frac{{t}^{2} + 1}{t} = 0$

#### Explanation:

L'Hospital's Rule can be used when we have a limit of and indeterminate form of $\frac{0}{0}$ (or by equally by inverting the form $\frac{\infty}{\infty}$).

So if ${\lim}_{x \rightarrow a} f \left(x\right) = 0$ $\mathmr{and} {\lim}_{x \rightarrow a} g \left(x\right) = 0$ , then

 lim_(x rarr a)f(x)/g(x) = lim_(x rarr a)(f′(x))/(g′(x))

For ${\lim}_{t \rightarrow 0} \ln \frac{{t}^{2} + 1}{t}$ we first need to check the conditions apply.

As $\ln 1 = 0$ hopefully you can see as $t \rightarrow 0$ both the numerator and denominator approach 0, so we can independently differentiate the numerator and denominator and apply L'Hospital's Rule

So:

${\lim}_{t \rightarrow 0} \ln \frac{{t}^{2} + 1}{t} = {\lim}_{t \rightarrow 0} \frac{\frac{d}{\mathrm{dt}} \left(\ln \left({t}^{2} + 1\right)\right)}{\frac{d}{\mathrm{dt}} \left(t\right)}$
$\text{ } = {\lim}_{t \rightarrow 0} \frac{\frac{1}{{t}^{2} + 1} \cdot 2 t}{1}$
$\text{ } = {\lim}_{t \rightarrow 0} \frac{2 t}{{t}^{2} + 1}$
$\text{ } = 0$

Which we can confirm graphically:
graph{ln(x^2+1)/x [-10, 10, -5, 5]}