Question

Limit problem-http:
//i.imgur.com/LVkiesQ.png
Help.

Answer 1

o.O? can you type it here please?

Answer 2

lim x->0 [sqrt(ax+b)-1]/x=1 Find the value of a and b.

Answer 3

\[lim_ {x\rightarrow 0} \frac {(ax+b)^{1/2}-1}{x}=1\] Correct?

Answer 4

Sorry it will be lim x->0 [sqrt(ax+b)-2]/x=1

Answer 5

Just 2 in place of 1.

Answer 6

\[lim_ {x\rightarrow 0} \frac {(ax+b)^{1/2}-2}{x}=1\]

Answer 7

Absolutely.

Answer 8

k so, let's get rid of that square root somehow

Answer 9

any ideas?

Answer 10

or, put everything in a square root

Answer 11

I cannot solve it by L'Hospital.Is there is any other way.....

Answer 12

no, it's undef 2/0 so correct no l'hops

Answer 13

so we need to mess with it algabraicly

Answer 14

Yes,but the problem is the x in the denominator.....

Answer 15

why is that a problem?

Answer 16

Here is a smart approach: Do you know the binomial expansion?

Answer 17

^^Yes that would be a nice approach, but I'm gonna try and do it without just in case they don't know it :) ^

We are given the limit. \[lim_ {x\rightarrow 0} \frac {(ax+b)^{1/2}-2}{x}=1\] right? so now, given that it =1 we know, that it must simplify to______.

Answer 18

It must simplify to 1.

Answer 19

right, so that means through some sort of manipulation, we can make it x/x or something else equivalent to 1

Answer 20

So, let's think

Answer 21

Yes

Answer 22

could a substitution or multiplying by the conjugate help us?

Answer 23

Yes,i did it but what about the x in the denominator.

Answer 24

That doesn't matter exactly just hold on

Answer 25

which did you try?

Answer 26

@AravindG yes i know bino.....

Answer 27

@FibonacciChick666 i tried it many times, but cannot do.....

Answer 28

no i mean which did you try, substitution or conjugate?

Answer 29

conjugate

Answer 30

The numerator is coming (ax+b-4).

Answer 31

ok, so how about this, what if you wanter to use l'hops, what would b have to be?

Answer 32

and try this sub \(u=(ax+b)^2\)

Answer 33

Where to substitute this.....

Answer 34

Where would it be useful?

Answer 35

But there is no (ax+b)^2,there is only (ax+b)^1/2.

Answer 36

(u^2)^1/2 ring a bell?

Answer 37

I coulnot understand u.....

Answer 38

\[(u^2)^{1/2}\]... think about it

Answer 39

a,b constant?

Answer 40

find values of a, b @mustafa2014

Answer 41

aha :)

Answer 42

Yes, a and b is constant.

Answer 43

did you figure out what I mean for the substitution yet?

Answer 44

If i try u=(ax+b)^1/2

Answer 45

you could try that too, but I just don't like having square roots

Answer 46

sorry, was II dyslexic? h/o no wonder you're confused

Answer 47

use u^2=ax+b

Answer 48

I'm sorry, about that

Answer 49

just remember to fix your limit and replace the x correctly

Answer 50

The problem is that there is no (ax+b),but there is (ax+b)^1/2

Answer 51

that isn't a problem...

Answer 52

You just put it in

Answer 53

try u=x+1

Answer 54

start by simplifying the expression
((ax+b)^(1/2)-2)/x=1
(ax+b)^1/2-2=x
(ax+b)^1/2=2+x
ax+b=(2+x)^2
ax+b=4+4x+x^2
x^2+(4-a)x+(4-b)=0
now that it is a quadratic in x, consider the limit.
as x->0, we have 4-b=0, thus b=4
plugging in b=4:
x^2+(4-a)x=0
x+(4-a)=0
as x->0, we have 4-a=0, thus a=4
thus a=4 and b=4

Answer 55

ok(this is against my principles,but im tired), you get this limit:
\[lim_{u\rightarrow b} \frac{u-2}{\frac{u^2-b}{a}}=1\]

Answer 56

Yes,i got this just a minute before.....

Answer 57

then just solve

Answer 58

just listen to @eashy and be done with this, their method is a lot more simple than what I am thinking of