# Thread: help me find this limit

1. ## help me find this limit

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

2. Originally Posted by silencecloak
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
What is the value that $\displaystyle t$ is approaching?

--Chris

3. Originally Posted by Chris L T521
What is the value that $\displaystyle t$ is approaching?

--Chris
Well I thought, nothing

But the site says I'm wrong

I don't think I answered your question, but I don't know how to

4. Originally Posted by silencecloak
Well I thought, nothing

But the site says I'm wrong

I don't think I answered your question, but I don't know how to
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.

5. Originally Posted by mr fantastic
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

6. Originally Posted by silencecloak
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
$\displaystyle \frac{1 - \sqrt{1+t}}{t \sqrt{1 + t}}$

$\displaystyle = \frac{(1 - \sqrt{1+t})}{t \sqrt{1 + t}} \times \frac{(1 + \sqrt{1 + t}) }{(1 + \sqrt{1 + t)}}$

$\displaystyle = \frac{(1 - \sqrt{1+t}) (1 + \sqrt{1 + t}) }{t \sqrt{1 + t} (1 + \sqrt{1 + t}) }$

$\displaystyle = \frac{-t}{t \sqrt{1 + t} (1 + \sqrt{1 + t}) }$.

Originally Posted by mr fantastic
[snip]
Multiplying this expression by $\displaystyle 1 = \frac{1 + \sqrt{1 + t} }{1 + \sqrt{1 + t}}$, Mr F says: Done.

simplify the numerator, Mr F says: Done.

cancel the common factor of t and then try taking the limit again. Mr F says: Left for you to do.

7. Originally Posted by mr fantastic
$\displaystyle \frac{1 - \sqrt{1+t}}{t \sqrt{1 + t}}$

$\displaystyle = \frac{(1 - \sqrt{1+t})}{t \sqrt{1 + t}} \times \frac{(1 + \sqrt{1 + t}) }{(1 + \sqrt{1 + t)}}$

$\displaystyle = \frac{(1 - \sqrt{1+t}) (1 + \sqrt{1 + t}) }{t \sqrt{1 + t} (1 + \sqrt{1 + t}) }$

$\displaystyle = \frac{-t}{t \sqrt{1 + t} (1 + \sqrt{1 + t}) }$.
Thank you for helping me