# Thread: Evaluate the given limit, Help

1. ## Evaluate the given limit, Help

Evaluate the Given Limit, this doesn't look complicated but I'm Stuck anyways how can i finish this 2 problems?

/ means over the other function
-> is an arrow as x approaches....

Lim x^2 - 16 / over sqr. root of X - 2
x->4

So far I have:
(X-4) (X+4) / sqr. root of 4 - 2 then i don't know if its right or how to finish it.

------------------------------------------------------------------------------

Lim sqr. root of 3 - t minus sqr. root of 3 / over t
t->0

I tried to multiply both sides by sqr. root of 3 - t plus sqr. root of 3

then i got
sqr. root of 3 - t / over sqr. root of 3 - t plus sqr. root of 3

I don't know how to finish or if i even started right.

If you can, can you please show the steps.

2. Originally Posted by Cyberman86
Evaluate the Given Limit, this doesn't look complicated but I'm Stuck anyways how can i finish this 2 problems?

/ means over the other function
-> is an arrow as x approaches....

Lim x^2 - 16 / over sqr. root of X - 2
x->4

So far I have:
(X-4) (X+4) / sqr. root of 4 - 2 then i don't know if its right or how to finish it.

------------------------------------------------------------------------------

Lim sqr. root of 3 - t minus sqr. root of 3 / over t
t->0

I tried to multiply both sides by sqr. root of 3 - t plus sqr. root of 3

then i got
sqr. root of 3 - t / over sqr. root of 3 - t plus sqr. root of 3

I don't know how to finish or if i even started right.

If you can, can you please show the steps.
use $x^{2}-16=(x-4)(x+4)=(\sqrt{x}-2)(\sqrt{x}+2)(x+4)$ for the first one, I dont understand the second one... why dont you try using paranthesis?

3. Originally Posted by Cyberman86
Evaluate the Given Limit, this doesn't look complicated but I'm Stuck anyways how can i finish this 2 problems?

/ means over the other function
-> is an arrow as x approaches....

[snip]

Lim sqr. root of 3 - t minus sqr. root of 3 / over t
t->0

I tried to multiply both sides by sqr. root of 3 - t plus sqr. root of 3

then i got
sqr. root of 3 - t / over sqr. root of 3 - t plus sqr. root of 3

I don't know how to finish or if i even started right.

If you can, can you please show the steps.
Is this what you mean? $\lim_{t\to{0}}\frac{\sqrt{3-t}-\sqrt{3}}{t}$?

--Chris

4. yes that's the exact problem but i don't how to write like that in here.

5. Hello, Cyberman86!

$\lim_{x\to4} \frac{x^2 - 16}{\sqrt{x} - 2}$

Factor twice: . $\lim_{x\to4} \frac{x^2-16}{\sqrt{x}-2} \;=\;\lim_{x\to4}\frac{(x-4)(x+4)}{\sqrt{x} - 2} \;=\; \lim_{x\to4}\frac{(\sqrt{x}-2)(\sqrt{x}+2)(x + 4)}{\sqrt{x}-2}$

Then: . $\lim_{x\to4}\,(\sqrt{x}+2)(x+4) \;=\;(\sqrt{4}+2)(4+4) \;=\;32$

$\lim_{t\to0}\frac{\sqrt{3-t} - \sqrt{3}}{t}$

Multiply top and bottom by $(\sqrt{3-t} + \sqrt{3})$

$\frac{\sqrt{3-t} - \sqrt{3}}{t}\cdot{\color{blue}\frac{\sqrt{3-t} + \sqrt{3}}{\sqrt{3-t} + \sqrt{3}}} \;=\;\frac{(3-t) - 3}{t(\sqrt{3-t} + \sqrt{3})} \;=$ . $\frac{-{\color{red}\rlap{/}}t}{{\color{red}\rlap{/}}t(\sqrt{3-t} + \sqrt{3})} \;=\;\frac{-1}{\sqrt{3-t} + \sqrt{3}}$

Then: . $\lim_{x\to0}\frac{-1}{\sqrt{3-t} + \sqrt{3}} \;=\; \frac{-1}{\sqrt{3} + \sqrt{3}} \;=\;\frac{-1}{2\sqrt{3}}$

6. Hello,

Another way for the second one.

Recall the formula : $\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}=f'(a)$

Here, $f(x)=\sqrt{x}$ and a=3.
$f'(x)=\frac 1{2 \sqrt{x}}$

$\lim_{t \to 0} \frac{\sqrt{3-t}-\sqrt{3}}{t}=\lim_{h \to 0} - \ \frac{\sqrt{3+h}-\sqrt{3}}{h}$, by substituting $h=-t$

$=- \ \frac{f(3+h)-f(3)}{h}=-f'(3)=- \ \frac{1}{2 \sqrt{3}}$