calculating limits question

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i've been working on this problem for an hour...
lim: h approaches 0 h^2/ (sq rt h^2 + h + 3) - (sq rt h + 3).

substitution doesn't work
i'm trying to multiply the denom by the reciprocal, but i'm just making a mess of it. converting the sq rt's to ( )^2 and then expanding that notation.
any help would be appreciated-
 
calc99 said:
lim: h approaches 0 h^2/ (sq rt h^2 + h + 3) - (sq rt h + 3).
Click to expand...
Do you mean the following?

. . . . .\(\displaystyle \L \begin{array}{c}limit\\h\rightarrow \0 \end{array}\,\, \frac{h^2}{\sqrt{h^2}\,+\,h\,+\,3}\,- \,\left(\sqrt{h}\,+\,3\right)\)

Please reply with confirmation or correction. Thank you.

Eliz.
 
no. the h + 3 is also in the denominator. also, both the h^2 + h + 3 and.... h + 3 are under square roots (all numbers & letters included).
 
So the limit is as follows...?

. . . . .\(\displaystyle \L \begin{array}{c}limit\\h\rightarrow \0 \end{array}\,\, \frac{h^2}{\sqrt{h^2\,+\,h\,+\,3}\,- \,\sqrt{h\,+\,3}}\)

Thank you.

Eliz.
 
Multiply top and bottom by the conjugate of \(\displaystyle \sqrt{h^{2}+h+3}-\sqrt{h+3}\)

It will simplify nicely, especially in the denominator.
 
so the denominator ends up being... h^2+h+3 minus sq rt h+3. orr. (my algebra frightens a lot of people)...(h^2 + h +3)^2 minus (h+3)^2..then broken down to... )...(h^2 + h +3) (h^2 + h +3) minus (h+3) (h+3)...
 
\(\displaystyle \L\\\frac{h^{2}(\sqrt{h^{2}+h+3}+\sqrt{h+3})}{(\sqrt{h^{2}+h+3}-\sqrt{h+3})\cdot(\sqrt{h^{2}+h+3}+\sqrt{h+3})}\)

I know, it looks menacing, but the denominator simplifies down to something very nice.
 
is h^4 +2h^3 + 6h^2 +9h +27 a good sign as to moving in the right direction or... is -5h-6 any better as to the denominator...
 
so.. = h^2 (sqrt h^2 + h + 3 plus sqrt h + 3) divided over (sq rt h^2 + h + 3)^2 minus (sq rt h + 3)^2)...
 
If it helps to FOIL, do it to this:

\(\displaystyle \L\\(\sqrt{h^{2}+h+3}-\sqrt{h+3})\cdot(\sqrt{h^{2}+h+3}+\sqrt{h+3})\)
 
so i'm left w/ sq rt h^2 +h +3 plus sq rt h+3 in the numerator, the denominator is 1...yees???
 
No, sorry. :cry:


\(\displaystyle \L\\(\sqrt{h^{2}+h+3}-\sqrt{h+3})\cdot(\sqrt{h^{2}+h+3}+\sqrt{h+3})=h^{2}\)

Maybe go thorugh the above until you see that it is h^2.

Try letting \(\displaystyle a=\sqrt{h^{2}+h+3}\;\ and\;\ b=\sqrt{h+3}\)

You'd have \(\displaystyle (a-b)(a+b)\)

That works out to \(\displaystyle a^{2}-b^{2}\)

Sub back in:

\(\displaystyle \L\\(\sqrt{h^{2}+h+3})^{2}-(\sqrt{h+3})^{2}\)

\(\displaystyle \L\\(h^{2}+h+3)-(h+3)=h^{2}+h+3-h-3=\overbrace{h^{2}}^{\text{Voila!}}\)

See?.

Remember, a square root times a square root cancels the radical and you're left with what's inside.

So, you have:

\(\displaystyle \L\\\lim_{h\to\0}\frac{h^{2}(\sqrt{h^{2}+h+3}+\sqrt{h+3})}{h^{2}}\)

Now, cancel what needs cancelled and you should see the limit.

See it?.
 
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