1. ## Limits question

Any chance you can help me on this one mate?

lim t -> 4 (4-t)/(5-[sqrt t^2 + 9])

i subbed in x = 4 and solved and got 0. right?

2. ## Re: Limits question

Both the numerator and denominator goes to 0 as t goes to 4, so you can't just plug in $\displaystyle t = 4$. L'Hospitals rule works here though, so I suggest you use it.

3. ## Re: Limits question

When t=4 you don't get 0 you get 0/0 which isn't defined.

4. ## Re: Limits question

Originally Posted by Hooperoo
Any chance you can help me on this one mate?

lim t -> 4 (4-t)/(5-[sqrt t^2 + 9])

i subbed in x = 4 and solved and got 0. right?
You can also do the following:
$\displaystyle \lim_{t \to 4} \frac{4-t}{5-\sqrt{t^2+9}} = \lim_{t \to 4} \frac{(4-t)(5+\sqrt{t^2+9})}{(5-\sqrt{t^2+9})(5+\sqrt{t^2+9})}$
$\displaystyle =\lim_{t \to 4} \frac{(4-t)(5+\sqrt{t^2+9})}{25-(t^2+9)}$
$\displaystyle =\lim_{t \to 4} \frac{(4-t)(5+\sqrt{t^2+9})}{16-t^2}$

Proceed ...

5. ## Re: Limits question

Originally Posted by Siron
You can also do the following:
$\displaystyle \lim_{t \to 4} \frac{4-t}{5-\sqrt{t^2+9}} = \lim_{t \to 4} \frac{(t-4)(5+\sqrt{t^2+9})}{(5-\sqrt{t^2+9})(5+\sqrt{t^2+9})}$
$\displaystyle =\lim_{t \to 4} \frac{(t-4)(5+\sqrt{t^2+9})}{25-(t^2+9)}$
$\displaystyle =\lim_{t \to 4} \frac{(t-4)(5+\sqrt{t^2+9})}{16-t^2}$

Proceed ...
Why has 4-t changed to t-4?

6. ## Re: Limits question

Originally Posted by biffboy
Why has 4-t changed to t-4?
It was a typo, thanks for noticing that!

7. ## Re: Limits question

Originally Posted by Siron
You can also do the following:
$\displaystyle \lim_{t \to 4} \frac{4-t}{5-\sqrt{t^2+9}} = \lim_{t \to 4} \frac{(4-t)(5+\sqrt{t^2+9})}{(5-\sqrt{t^2+9})(5+\sqrt{t^2+9})}$
$\displaystyle =\lim_{t \to 4} \frac{(4-t)(5+\sqrt{t^2+9})}{25-(t^2+9)}$
$\displaystyle =\lim_{t \to 4} \frac{(4-t)(5+\sqrt{t^2+9})}{16-t^2}$

Proceed ...
Essentially, cant divide by 0, so limit is not defined?

8. ## Re: Limits question

You can factore $\displaystyle 16-t^2=(4-t)(4+t)$

9. ## Re: Limits question

Originally Posted by Siron
You can factore $\displaystyle 16-t^2=(4-t)(4+t)$
so im getting lim t -> 4 (4-t)(5-sqrt t^2+9)/(4-t)(4+t)

do i then just cancel the (4-t) and solve like that?

Indeed.

11. ## Re: Limits question

$x_0=4$ is is called a removable discontinuity point.