Solving Limit of Question with Given Solution -3/2

[seal308]

Hi,

I need help with the following limit, the solution is apparently -3/2 but I don't get it.

Question:
limit as n approaches infinity of [ sqrt(n^2 +3n - 4) - sqrt(n^2 + 6n +5) ]

Attempt: So I was just thinking to factor out n

like: (n^2)^(1/2) (1 + 3/n - 4/n^2)^1/2 - n (n^2)^(1/2) (1 + 6/n + 5/n^2)^1/2

(n^2)^(1/2) simplifies to just n. factor n out from both terms.
Also 3/n, 4/n^2, 6/n, 5/n^2 all go to 0.

So when you clear things up: n [ 1^1/2 - 1^1/2] so that's a limit of infinity times 0, so I put the n in the denominator and then use l'hoptial.

[ 1^1/2 - 1^1/2] / n^-1
top is just a constant to derivative of that is just 0.

That is as far as I got.

 

[Jameson]

Hi seal308,

Usually with these kinds of problems it's not "allowed" to use L'Hopital's Rule or it's strongly discouraged. Let's see if we can find an algebraic solution. :)

\(\displaystyle \lim_{n \rightarrow \infty}\sqrt{n^2 +3n - 4} - \sqrt{n^2 + 6n +5}\)

Let's multiply by the conjugate of this expression: \(\displaystyle \frac{\sqrt{n^2 +3n - 4} +\sqrt{n^2 + 6n +5}}{\sqrt{n^2 +3n - 4} +\sqrt{n^2 + 6n +5}}\)

\(\displaystyle \sqrt{n^2 +3n - 4} - \sqrt{n^2 + 6n +5} \left(\frac{\sqrt{n^2 +3n - 4} +\sqrt{n^2 + 6n +5}}{\sqrt{n^2 +3n - 4} +\sqrt{n^2 + 6n +5}} \right) = \frac{n^2+3n-4-(n^2+6n+5)}{\sqrt{n^2 +3n - 4} +\sqrt{n^2 + 6n +5}}\)

What do we get when we simplify the numerator? Any idea what to do next?

 

[seal308]

Hi seal308,

Usually with these kinds of problems it's not "allowed" to use L'Hopital's Rule or it's strongly discouraged. Let's see if we can find an algebraic solution. :)

\(\displaystyle \lim_{n \rightarrow \infty}\sqrt{n^2 +3n - 4} - \sqrt{n^2 + 6n +5}\)

Let's multiply by the conjugate of this expression: \(\displaystyle \frac{\sqrt{n^2 +3n - 4} +\sqrt{n^2 + 6n +5}}{\sqrt{n^2 +3n - 4} +\sqrt{n^2 + 6n +5}}\)

\(\displaystyle \sqrt{n^2 +3n - 4} - \sqrt{n^2 + 6n +5} \left(\frac{\sqrt{n^2 +3n - 4} +\sqrt{n^2 + 6n +5}}{\sqrt{n^2 +3n - 4} +\sqrt{n^2 + 6n +5}} \right) = \frac{n^2+3n-4-(n^2+6n+5)}{\sqrt{n^2 +3n - 4} +\sqrt{n^2 + 6n +5}}\)

What do we get when we simplify the numerator? Any idea what to do next?

I get it!
Simplify denom is 2n so do limit as n approaches infinity of (-3n - 9) /(2n)
Factor out the n: limit as n approaches infinity of n(-3 - 9/n) /n(2)
n cancels out, 9/n goes to 0 so you get -3/2.

But how do you know not to use l'hopitals rule and multiply by conjugate.
Was it because it involved square roots?

 

[Jameson]

I agree that the numerator simplifies to $-3n-9$, but I'm not sure how you got the denominator reduces to $2n$. Could you show your work?

What I would do is factor out an $n^2$ from each term within the two square-roots.

For example: \(\displaystyle \sqrt{n^2 +3n - 4} = \sqrt{n^2\left(1+\frac{3}{n}-\frac{4}{n^2}\right)}=\sqrt{n^2} \sqrt{1+\frac{3}{n}-\frac{4}{n^2}}=n\sqrt{1+\frac{3}{n}-\frac{4}{n^2}} \)

I'll let you try the other one. To answer your question about how I knew to do this - the short answer is lots of practice. Once you do a lot of these limit problems you learn the tricks you have to use and multiplying by the conjugate is a very common one. If you see a limit in this form with a radical minus another radical as $x \rightarrow \infty$, then it might be a good thing to try. If you have a fraction with radicals in it and directly plugging in the limit gives you issues, then the conjugate might work then too. :)

 

[seal308]

I agree that the numerator simplifies to $-3n-9$, but I'm not sure how you got the denominator reduces to $2n$. Could you show your work?

What I would do is factor out an $n^2$ from each term within the two square-roots.

For example: \(\displaystyle \sqrt{n^2 +3n - 4} = \sqrt{n^2\left(1+\frac{3}{n}-\frac{4}{n^2}\right)}=\sqrt{n^2} \sqrt{1+\frac{3}{n}-\frac{4}{n^2}}=n\sqrt{1+\frac{3}{n}-\frac{4}{n^2}} \)

I'll let you try the other one. To answer your question about how I knew to do this - the short answer is lots of practice. Once you do a lot of these limit problems you learn the tricks you have to use and multiplying by the conjugate is a very common one. If you see a limit in this form with a radical minus another radical as $x \rightarrow \infty$, then it might be a good thing to try. If you have a fraction with radicals in it and directly plugging in the limit gives you issues, then the conjugate might work then too. :)

I similified the denominator as follows:

sqrt(n^2 + 3n -4) + sqrt(n^2 + 6n +5)
which becomes (n^2)^(1/2) (1 + 3/n + 4/n^2)^1/2 + (n^2)^(1/2) (1+6/n + 5/n)^1/2
(n^2)^(1/2) simplifies to just n, factor n out. and 3/n, 4/n^2, 6/n and 5/n all go to 0 as n approaches infinity so i just said they go to 0.

So clearing thigs up you have in the denominator n (1^1/2+1^1/2) which is just n (1+1)

Is that wrong?

 

[Jameson]

Ok, I see what you mean now. You applied the limit before canceling when you wrote the denominator becomes $2n$, but I see what you meant now.. I applied the limit at the end. Your algebra is correct but I would wait to apply the limit. Yep, I think you've got it!

 

[alyafey22]

If you are a fan of differentiation then by substituting $k=n^{-1}$

$$\lim_{k \rightarrow 0} \frac{\sqrt{1 +3k - k^2} - \sqrt{1 + 6k +5k^2}}{k} = \lim_{k \rightarrow 0} \frac{f(k)-f(0)}{k-0} = f'(0) $$

 

[MarkFL]

Hi,

I need help with the following limit, the solution is apparently -3/2 but I don't get it.

Question:
limit as n approaches infinity of [ sqrt(n^2 +3n - 4) - sqrt(n^2 + 6n +5) ]

Attempt: So I was just thinking to factor out n

like: (n^2)^(1/2) (1 + 3/n - 4/n^2)^1/2 - n (n^2)^(1/2) (1 + 6/n + 5/n^2)^1/2

(n^2)^(1/2) simplifies to just n. factor n out from both terms.
Also 3/n, 4/n^2, 6/n, 5/n^2 all go to 0.

So when you clear things up: n [ 1^1/2 - 1^1/2] so that's a limit of infinity times 0, so I put the n in the denominator and then use l'hoptial.

[ 1^1/2 - 1^1/2] / n^-1
top is just a constant to derivative of that is just 0.

That is as far as I got.

If I were going to apply L'Hôpital's Rule here, I would first write the limit as:

\(\displaystyle L=\lim_{n\to\infty}\frac{\sqrt{1+\dfrac{3}{n}-\dfrac{4}{n^2}}-\sqrt{1+\dfrac{6}{n}+\dfrac{5}{n^2}}}{\dfrac{1}{n}}\)

Now, for the sake of simplcity, let's use the substitution: \(\displaystyle u=\frac{1}{n}\) and we may write:

\(\displaystyle L=\lim_{u\to0}\frac{\sqrt{1+3u-4u^2}-\sqrt{1+6u+5u^2}}{u}\)

We have the indeterminate form 0/0, and so L'Hôpital's Rule allows:

\(\displaystyle L=\lim_{u\to0}\left(\frac{3-8u}{2\sqrt{1+3u-4u^2}}-\frac{5u+3}{\sqrt{1+6u+5u^2}}\right)=\frac{3}{2}-3=-\frac{3}{2}\)