How to Solve Limits Involving Square Roots and Infinity?

[askor]

How to find the

$\lim_{x \to \infty} (\sqrt{ax^2 + bx + c} - \sqrt{ax^2 + px + q}) = \frac{b - p}{2 \sqrt{a}}$

?

Here is what I get

$\lim_{x \to \infty} (\sqrt{ax^2 + bx + c} - \sqrt{ax^2 + px + q})$
$= \sqrt{\lim_{x \to \infty} ax^2 + \lim_{x \to \infty} bx + \lim_{x \to \infty} c} - \sqrt{\lim_{x \to \infty} ax^2 + \lim_{x \to \infty} px + \lim_{x \to \infty} q}$
$=\sqrt{\infty + \infty + c} - \sqrt{\infty + \infty + q}$
$= \sqrt{\infty} - \sqrt{\infty}$
$= \infty$

 

[Samy_A]

How to find the

$\lim_{x \to \infty} (\sqrt{ax^2 + bx + c} - \sqrt{ax^2 + px + q}) = \frac{b - p}{2 \sqrt{a}}$

?

Here is what I get

$\lim_{x \to \infty} (\sqrt{ax^2 + bx + c} - \sqrt{ax^2 + px + q})$
$= \sqrt{\lim_{x \to \infty} ax^2 + \lim_{x \to \infty} bx + \lim_{x \to \infty} c} - \sqrt{\lim_{x \to \infty} ax^2 + \lim_{x \to \infty} px + \lim_{x \to \infty} q}$
$=\sqrt{\infty + \infty + c} - \sqrt{\infty + \infty + q}$
$= \sqrt{\infty} - \sqrt{\infty}$
$= \infty$

∞-∞ is undefined, so the last step is wrong.

Hint: $(\sqrt y- \sqrt z)(\sqrt y+ \sqrt z)=y-z$

 

[WWGD]

OP: Or you can complete the square inside each of the radicals and ignore the constant terms.

 

[askor]

Hint:
$(\sqrt y- \sqrt z)(\sqrt y+ \sqrt z)=y-z$

Do you mean

$(\sqrt{ax^2 + bx + c} - \sqrt{ax^2 + px + q})(\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q})$
$= (ax^2 + bx + c)-(ax^2 + px + q)$

??

 

[Samy_A]

Do you mean

$(\sqrt{ax^2 + bx + c} - \sqrt{ax^2 + px + q})(\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q})$
$= (ax^2 + bx + c)-(ax^2 + px + q)$

??

Yes, that is correct (although not exactly what I meant).
You can further simplify the result.
After that, use this other mathematical identity: $A=\frac{A*B}{B}$, where $B\neq 0$.

 

[WWGD]

Do you mean

$(\sqrt{ax^2 + bx + c} - \sqrt{ax^2 + px + q})(\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q})$
$= (ax^2 + bx + c)-(ax^2 + px + q)$

??

No, he surely meant (sorry to speak for you Samy_A) that you must both multiply and divide by this expression , otherwise , if you only multiply and don't divide, you are changing the original expression for which you want to find the limit.

 

[askor]

CAn we just go to the point how to find the result? Please tell me step-by-step, I am confuse.

 

[Mark44]

CAn we just go to the point how to find the result? Please tell me step-by-step, I am confuse.

That's not how it works here. We will steer you in the right direction, but you need to do the majority of the work.

Do you mean

$(\sqrt{ax^2 + bx + c} - \sqrt{ax^2 + px + q})(\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q})$
$= (ax^2 + bx + c)-(ax^2 + px + q)$

Instead of multiplying by $(\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q})$, multiply by 1, in the form of $(\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q})$ over itself. Multiplying by 1 doesn't change the value of an expression, so is always valid to do.

 

[askor]

Instead of multiplying by $(\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q})$, multiply by 1, in the form of $(\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q})$ over itself. Multiplying by 1 doesn't change the value of an expression, so is always valid to do.

Isn't multplying by 1 (in the form of $(\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q})$) just make more complex because the denominator is in the form of square root?

$(\sqrt{ax^2 + bx + c} - \sqrt{ax^2 + pc + q}) × \frac{\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q}}{\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q}}$

$= \frac{(ax^2 + bx + c) - (ax^2 + px + q)}{\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q}}$
$=\frac{bx + c - px - q}{\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q}}$
$=\frac{bx - px + c - q}{\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q}}$
$=\frac{(b - p)x + c - q}{\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q}}$

 

[Samy_A]

Isn't multplying by 1 (in the form of $(\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q})$) just make more complex because the denominator is in the form of square root?

$(\sqrt{ax^2 + bx + c} - \sqrt{ax^2 + pc + q}) × \frac{\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q}}{\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q}}$

$= \frac{(ax^2 + bx + c) - (ax^2 + px + q)}{\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q}}$
$=\frac{bx + c - px - q}{\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q}}$
$=\frac{bx - px + c - q}{\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q}}$
$=\frac{(b - p)x + c - q}{\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q}}$

Ok, now look at what you have.
Let's first make an informal argument. As you need the limit for $x \to + \infty$, we can consider only the highest order terms in both numerator and denominator (because for large $x$ they are much larger than the other terms).
That gives $\frac{(b-p)x}{\sqrt{ax^2}+\sqrt{ax^2}}=\frac{(b-p)x}{2x\sqrt{a}}=\frac{b-p}{2\sqrt{a}}$.

Now all you have to do is make this informal argument formally correct.

 

[askor]

Ok, now look at what you have.
Let's first make an informal argument. As you need the limit for $x \to + \infty$, we can consider only the highest order terms in both numerator and denominator (because for large $x$ they are much larger than the other terms).
That gives $\frac{(b-p)x}{\sqrt{ax^2}+\sqrt{ax^2}}=\frac{(b-p)x}{2x\sqrt{a}}=\frac{b-p}{2\sqrt{a}}$.

Now all you have to do is make this informal argument formally correct.

How do you get $\frac{(b-p)x}{\sqrt{ax^2}+\sqrt{ax^2}}$? What happen to the c - q in the numerator and bx + c, and px + q in the denominator?

 

[Samy_A]

How do you get $\frac{(b-p)x}{\sqrt{ax^2}+\sqrt{ax^2}}$? What happen to the c - q in the numerator and bx + c, and px + q in the denominator?

It's explained in the previous post.

Take the numerator: $(b-p)x+(c-q)$. When $x$ gets very large, the value of the numerator will mainly be determined by $(b-p)x$. The $c-q$ term is constant, and will become negligible compared to $(b-p)x$ for large $x$. Similarly for the denominator.
That's the informal computation I did in the previous post, and you can see that the result is the searched limit.

Now convert this informal argument into a formal one. Hint: Divide both numerator and denominator by $x$, take the $1/x$ in the denominator into the square roots, and then take the limit.

 

[askor]

It's explained in the previous post.

Take the numerator: $(b-p)x+(c-q)$. When $x$ gets very large, the value of the numerator will mainly be determined by $(b-p)x$. The $c-q$ term is constant, and will become negligible compared to $(b-p)x$ for large $x$. Similarly for the denominator.
That's the informal computation I did in the previous post, and you can see that the result is the searched limit.

Now convert this informal argument into a formal one. Hint: Divide both numerator and denominator by $x$, take the $1/x$ in the denominator into the square roots, and then take the limit.

I still don't understand.

Btw, this is the calculus textbook I use to learn about limit:

<< Illegal download link deleted by Moderator >>

Can someone please tell me what page of above book about limit x approach infinity similar to above problem?

 

[Samy_A]

I still don't understand.

Divide both numerator and denominator by $x$:
$$ \frac{(b - p)x + c - q}{\sqrt{ax^2 + bx + c} + \sqrt{ax^2 + px + q}}=\frac{(b - p) + \frac{c - q}{x}}{\frac{1}{x}\sqrt{ax^2 + bx + c} + \frac{1}{x}\sqrt{ax^2 + px + q}}$$
Now take the $1/x$ in the denominator into the square roots, and then take the limit.

 

[Erland]

askor, the most similar example in your book is Example 9. p. 107, but it is a lot simpler than this one. Exercises 80-86, p. 116 are af the same type as your example. If you have a solution manual you might look up the solutions of these.