# Calculate $\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}$

$$\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}$$

Here I've tried multiplying by $\sqrt[4]{x+9}+2$ and few other method.

Thanks in advance for solution / hints using simple methods.

Edit

Please don't use l'Hosplital rule. We are before derivatives, don't know how to use it correctly yet. Thanks!

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Why does it seem no one can ever use L'Hôpital? Poor Guillaume... –  David Mitra Jan 11 '13 at 20:04
Because it's just too easy. He's not solving an important problem, he's practicing. Plus he hasn't learned it yet. –  Git Gud Jan 11 '13 at 20:05
The correct use is \lim_{limit} rather than {\underset ...} like you keep doing. I already corrected it in one or two posts and left one comment before. And another possibility is to use \lim\limits_{blah} (which have no effect in $$ environments by the way). – Asaf Karagila Jan 11 '13 at 20:13 Can you use Talyor expansion? – Patrick Li Jan 11 '13 at 20:31 He can, I saw it on another post. – Git Gud Jan 11 '13 at 20:36 ## 5 Answers One thing you should learn, is that analysts like to think of functions as power series (or at worst Laurent series) In this sense, L'Hopital's rules is essentially saying that "When we have a function  \frac {(x-7)f(x)}{(x-7)g(x)}, then we can 'fill in' the hole and carry along our own merry way". So, if we don't have L'Hopital, and we know we want to use it, we simply force it out. For example, notice that$$(\sqrt[4]{x+9} - 2)(\sqrt[4]{x+9} + 2)= \sqrt[2]{x+9} -4,$$which I'm sure you did. Does this help us? No, not yet, because we haven't forced out the troublesome x-7. So let's try again, and we use$$(\sqrt{x+9}-4)(\sqrt{x+9}+4) = x+9 - 16 = x-7.$$Are we done with the denominator? You bet! How about the numerator? It is likely giving us problems with x-7, so let's force it out. Try$$(\sqrt{x+2} - \sqrt[3]{x+20})(\sqrt{x+2} + \sqrt[3]{x+20}) = x+2 - (x+20)^{2/3}.$$Are we done? No not yet, I don't see a x-7. So let's use$$ [(x+2) - (x+20)^{2/3} ][(x+2)^2 + (x+2)(x+20)^{2/3} + (x+20^{4/3} ] = (x+2)^3 - (x+20)^2.$$Are we done? I most certainly hope so, and you can check that we can factor out an (x-7), since (7+2)^3 - (7+20)^2 = 0. What's the moral of the story?$$\frac {\sqrt{x+2} - \sqrt[3]{x+20}} {\sqrt[4]{x+9} - 2} \times \frac {\mbox{stuff}} {\mbox{same stuff}} = \frac {(x-7) \times \mbox {something}}{(x-7) \times \mbox {more something}}.$$And now we rejoice and wait for the cows to come home. - The finial answer. Nice hint for the OP. – Babak S. Jan 11 '13 at 20:39 Hint: Use that \frac{a^4-b^4}{a-b} = a^3 + a^2b + ab^2 + b^3. Setting a=\sqrt[4]{x+9} and b=2, you see a^4-b^4 = x-7, and you get$$\frac{1}{\sqrt[4]{x+9}-2} = \frac{a^3 + a^2b + ab^2 + b^3}{x-7}$$Similarly you can write \sqrt{x+2}-3 as:$$\sqrt{x+2}-3 = \frac{x-7}{\sqrt{x+2}+3}$$And a similar but uglier result for \sqrt[3]{x+20}-3 using u-v=\frac{u^3-v^3}{u^2+uv+v^2} with u=\sqrt[3]{x+20} and v=3, that gives u^3-v^3 = x-7. Note then that the x-7s cancel out, and you get an expression where none of the numerators or denominators approach zero as x\to 7, so you can finally just plug in x=7 in that expression. Cancelling out the x-7 terms, you get:$$(a^3 + a^2b + ab^2 + b^3)\left(\frac{1}{\sqrt{x+2}+3}-\frac{1}{u^2+uv+v^2}\right)$$But as x\to 7 a\to b=2 and u\to v=3. So the limit is:$$(4\cdot2^3)\left(\frac{1}{6}-\frac{1}{3\cdot 3^2}\right)$$- You could use Taylor series:$$\lim_{x\to 7}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2} = \lim_{x\to 7}\frac{\sqrt{x-7+9}-\sqrt[3]{x-7+27}}{\sqrt[4]{x-7+16}-2}$$Now let y=x-7, we have$$\lim_{y\to 0}\frac{(y+9)^{1/2}-(y+27)^{1/3}}{(y+16)^{1/4}-2}\\ =\frac{3}{2} \lim_{y\to 0}\frac{(1+\frac{y}{9})^{1/2}-(1+\frac{y}{27})^{1/3}}{(1+\frac{y}{16})^{1/4}-1}$$Now you can Taylor expand it and keep the first order terms, and the final result will follow. - #$$\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}$$#$$=\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{{(x+9)^\frac{1}{4}}-(16)^\frac{1}{4}}.\frac{x-7}{x-7}$$#$$=\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{x-7}.\underset{x\rightarrow7}{\lim}\frac{x-7}{{(x+9)^\frac{1}{4}}-(16)^\frac{1}{4}}$$#$$=\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}-3+3}{x-7}.\underset{x\rightarrow7}{\lim}\frac{x+9-16}{{(x+9)^\frac{1}{4}}-(16)^\frac{1}{4}}$$#$$=[\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt9}{x-7}-\underset{x\rightarrow7}{\lim}\frac{\sqrt[3]{x+20}-\sqrt[3]27}{x-7}].\underset{x\rightarrow7}{\lim}\frac{(x+9)-16}{{(x+9)^\frac{1}{4}}-(16)^\frac{1}{4}}$$#$$= \frac{112}{27}

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How did you do the last step? I followed it all the way until the $=\frac{122}{27}$ –  Joao Aug 30 at 5:30

112/27 which is roughly 4.14815

the derivative of the top at x=7 is 7/54

the derivative of the bottom at x=7 is 1/32

(7/54)/(1/32) is 112/27

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Please tell me how, because I've got 0/0 –  Steve Jan 11 '13 at 20:01
when you have 0/0 the limit of that formula is equal to the derivative of the top divided by the derivative of the bottom at that x. that is l'hopital. Is this for calculus class? –  kaine Jan 11 '13 at 20:12
I'm trainig before exam. currently we are before derivatives –  Steve Jan 11 '13 at 20:15
calvin lin's answer is correct, disregard mine until you learn calculus –  kaine Jan 11 '13 at 20:54
I think, Steve needs more time till he learns derivatives and its application. So the best answers here are two other ones. :-) –  Babak S. Jan 11 '13 at 21:03