Question

How do I rewrite the limit part in terms of the substitution.

Answer 1

I need to know how to write the part for the approach section. ex \[\lim_{x \rightarrow a} \frac{\sqrt{x}-a}{x-a}\]

If I sub root (x) with u how do I write \[\lim_{u \rightarrow ?}\]

Answer 2

thats ax+b, right ? and x->0
let y= ax+b , when x->0 , y-> a(0)+b, y->b

Answer 3

O alright ok. Thanks so I just make that y = a() + b so y approc b

Answer 4

\[m_{y \rightarrow b} \frac{(\sqrt{y}-2)a}{u-b}\]

now I am even more stuck

Answer 5

you mean y-b in the denom, right ?

Answer 6

yes

Answer 7

The step you did here I wouldn't even have done.
The limit exist when you have 0/0, right?
So to find what b is set x=0.

Answer 8

In the numerator.

Answer 9

how do I do that?

Answer 10

sqrt(a(0)+b)-2=0

I set the top equal to 0 and set x equal to 0 because x approaches 0.

Answer 11

After you find b, try rationalizing the top for your journey to find a.

Answer 12

\[\lim _{\xrightarrow 0} \frac{\sqrt{ax-b}-2}{x} =1 \]

wait wait wait wait.... you just did what. I aready changed the number stuff. first

if I write the limit in terms of y.. y approaches b as x aproches0

Answer 13

The limit exist when you have 0/0.

Answer 14

mebs, listen to myin, you can get b easily without doing any of that

Answer 15

so then approaches x

Answer 16

x approaches 0

Answer 17

o ok

Answer 18

\[\sqrt{a(0)+b}-2=0\]

Answer 19

I want to know when the top is 0 for when x is 0.
I want the top to be zero because I want 0/0.
I put x as 0 because x approaches 0.

Answer 20

so \[b = 2^{2}\]

Answer 21

Yep.

Answer 22

But that's cheating. You can't do that if it is supposed to be 0/0 even though you know it equals 1 .

Answer 23

Now look at your whole problem where b is 4 and let's find a.
Hint: You will need to do some rationalizing.

Answer 24

How is that cheating?

Answer 25

because you just let an undefined value equal and let it equal 2 then squared it for b even though its undefined. 2/0

Answer 26

We know the limit exist so we know we must have 0/0.

Answer 27

We don't have 2/0

Answer 28

alright so I guess its clever.

Answer 29

It is 0/0.

Answer 30

the concept is,
since the limit existed, and the denominator =0
the denominator should also have a factor of that leads to 0
there's only one factor in numerator = \sqrt {...}-2
so that must =0
else if the num NOT =0, the limit won't exist.

Answer 31

So now I just evaluate \[\lim_{x \rightarrow 0} \frac{\sqrt{ax +4}-2}{x}\]

Answer 32

You would go about it like you are evaluating it now.
You would rationalize the top.

Answer 33

so like the above part or the substitution part?

Answer 34

Or use l'hospital
I wouldn't even use a sub.

Answer 35

I am not allowed to use l'hoptial ....I haven't learned it in class =(

Answer 36

rationalizing is easier than substitution

Answer 37

Yep then use rationalizing.

Answer 38

I'm not saying you can't use a sub. I'm just saying I wouldn't.

Answer 39

so just multiply the top and bottom by \[\frac{\sqrt{ax+4}+2}{\sqrt{ax+4}+2}\]

Answer 40

Yes.

Answer 41

no...

Answer 42

I meant that b = 4 and

Answer 43

well I get a/4

Answer 44

\[\lim_{x \rightarrow 0}\frac{\sqrt{ax+4}-2}{x} \cdot \frac{\sqrt{ax+4}+2}{\sqrt{ax+4}+2}=\lim_{x \rightarrow 0} \frac{(ax+4)-4}{x(\sqrt{ax+4}+2)}\]

Ok good/
a/4
So a/4=1

Answer 45

That wouldn't make a 1 or -1.

Answer 46

alright that was nice.. thanks

Answer 47

Great. I have seen that question a dozen times now on openstudy. lol.

Answer 48

Sometimes we did l'hospital and other times we did rationalizing.

Answer 49

Alright time to attempt similar problems... I hope you guys are still around until I get it haha thanks @hartnn and @myininaya