Question

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

Answer 1

`lim_(t rarr 0)ln(t^2+1)/t = 0`

Explanation:

L'Hospital's Rule can be used when we have a limit of and indeterminate form of `0/0` (or by equally by inverting the form `oo/oo`).

So if `lim_(x rarr a) f(x) = 0` `and lim_(x rarr a)g(x)=0` , then

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

For `lim_(t rarr 0)ln(t^2+1)/t` we first need to check the conditions apply.

As `ln1=0` hopefully you can see as `t rarr 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 rarr 0)ln(t^2+1)/t = lim_(t rarr 0) (d/dt(ln(t^2+1)))/(d/dt(t))`
`" " = lim_(t rarr 0) (1/(t^2+1)*2t)/1`
`" " = lim_(t rarr 0) (2t)/(t^2+1)`
`" " = 0`

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