Finding Constants To A Limit Problem

[Bashyboy]

Find all the values of the constants a and b such that
$\lim_{x\rightarrow0}\frac{\sqrt{a+bx} -\sqrt{3}}{x} = \sqrt{3}$

I tried to multiply by the conjugate, but it indeed failed to give me any insight as to how to solve this problem. Could someone possibly prod me into the right direction?

 

[tiny-tim]

Hi Bashyboy!

Start with a …

what must a be?​

(can a = 0? 2? …)

 

[Bashyboy]

So, in order to start this problem, we have to begin with speculation? There is no analytic way of solving for a? Well, I know it can't be zero, because then all you would have under the radical is zero.

 

[SammyS]

Find all the values of the constants a and b such that
$\lim_{x\rightarrow0}\frac{\sqrt{a+bx} -\sqrt{3}}{x} = \sqrt{3}$

I tried to multiply by the conjugate, but it indeed failed to give me any insight as to how to solve this problem. Could someone possibly prod me into the right direction?

Since the denominator → 0 , the only way for the limit to exist is for the numerator → 0 as x → 0. Right ?

 

[tiny-tim]

So, in order to start this problem, we have to begin with speculation?

No, but trying one or two values may help us to see what we need to avoid

 

[Noesis]

Note that this is an equation with a limit on one side. The limit, if it converges evaluates to a number. So this is a relationship between two numbers.

Does the statement as provided imply the validity of any further statements? Equivalent ones that can be manipulated and better understood to reach an answer?

Consider that the given equation implies:

$\frac{\lim_{x \to 0}\sqrt{a+bx}-\sqrt{3}}{\lim_{x \to 0}x} = \sqrt{3}$

which implies:

$\lim_{x \to 0}\sqrt{a+bx}-\sqrt{3} = \lim_{x \to 0}\sqrt{3}x$

From this, you should be able to figure out certain restrictions on the variables $a$ and $b$.

 

[SammyS]

Note that this is an equation with a limit on one side. The limit, if it converges evaluates to a number. So this is a relationship between two numbers.

It looks like it will be a valid two-sided limit when all is done.

Does the statement as provided imply the validity of any further statements? Equivalent ones that can be manipulated and better understood to reach an answer?

Consider that the given equation implies:

$\frac{\lim_{x \to 0}\sqrt{a+bx}-\sqrt{3}}{\lim_{x \to 0}x} = \sqrt{3}$

What you have written above is absolutely improper. The denominator you show is identically zero.

which implies:

$\lim_{x \to 0}\sqrt{a+bx}-\sqrt{3} = \lim_{x \to 0}\sqrt{3}x$

From this, you should be able to figure out certain restrictions on the variables $a$ and $b$.

 

[Bashyboy]

Since the denominator → 0 , the only way for the limit to exist is for the numerator → 0 as x → 0. Right ?

Okay, so I set the numerator equal to zero, and took the limit of it, and I got a = 3; but now I am not sure where to plug in a = 3 so I can solve for b.

 

[Infinitum]

Okay, so I set the numerator equal to zero, and took the limit of it, and I got a = 3; but now I am not sure where to plug in a = 3 so I can solve for b.

Since you have both your numerator and denominator approaching zero, that means you have a 0/0 indeterminate limit form. How do you solve such limits?

 

[SammyS]

Okay, so I set the numerator equal to zero, and took the limit of it, and I got a = 3; but now I am not sure where to plug in a = 3 so I can solve for b.

Now that you have a =3, look at your result from multiplying by the conjugate. That should tell you what to use for b.

 

[Bashyboy]

Does it sound right that b = 6?

 

[Infinitum]

Does it sound right that b = 6?

Yep! That is correct

 

[Bashyboy]

Thank you all very much, all of you provided great insight.