im pretty sure im getting the right answer that there is no limit
but the online math problem for my math class says i am wrong
$\displaystyle \frac{1}{t\sqrt{1+t}}-\frac{1}{t}$
thanks
If you're being asked to find a limit then you surely have been told the value that t is approaching?? Is this what your're trying to find:
$\displaystyle \lim_{t \rightarrow 0} \left( \frac{1}{t\sqrt{1+t}}-\frac{1}{t}\right)$.
The value of the limit is $\displaystyle - \frac{1}{2}$.
Note that $\displaystyle \frac{1}{t\sqrt{1+t}}-\frac{1}{t} = \frac{1 - \sqrt{1+t}}{t \sqrt{1 + t}}$.
Multiplying this expression by $\displaystyle 1 = \frac{1 + \sqrt{1 + t} }{1 + \sqrt{1 + t}}$, simplify the numerator, cancel the common factor of t and then try taking the limit again.
oh, im sorry Chris i misunderstood your question
and I still don't see where you are getting 1/2 from
I always end up with $\displaystyle t(t+1)$ in the denominator which when u plug in 0....is 0
$\displaystyle \frac{1}{t\sqrt{1+t}}-\frac{1}{t} = \frac{x - \sqrt{1+t}}{(x)t \sqrt{1 + t}} $
you can cancel those two "x" out? seems weird to me
$\displaystyle \frac{1 - \sqrt{1+t}}{t \sqrt{1 + t}}$
$\displaystyle = \frac{(1 - \sqrt{1+t})}{t \sqrt{1 + t}} \times \frac{(1 + \sqrt{1 + t}) }{(1 + \sqrt{1 + t)}}$
$\displaystyle = \frac{(1 - \sqrt{1+t}) (1 + \sqrt{1 + t}) }{t \sqrt{1 + t} (1 + \sqrt{1 + t}) }$
$\displaystyle = \frac{-t}{t \sqrt{1 + t} (1 + \sqrt{1 + t}) }$.