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

Solution 1:

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.

Solution 2:

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

$$=\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{{(x+9)^\frac{1}{4}}-(16)^\frac{1}{4}}.\frac{x-7}{x-7}$$

$$=\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{x-7}.\underset{x\rightarrow7}{\lim}\frac{x-7}{{(x+9)^\frac{1}{4}}-(16)^\frac{1}{4}}$$

$$=\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt[3]{x+20}-3+3}{x-7}.\underset{x\rightarrow7}{\lim}\frac{x+9-16}{{(x+9)^\frac{1}{4}}-(16)^\frac{1}{4}}$$

$$=[\underset{x\rightarrow7}{\lim}\frac{\sqrt{x+2}-\sqrt9}{x-7}-\underset{x\rightarrow7}{\lim}\frac{\sqrt[3]{x+20}-\sqrt[3]27}{x-7}].\underset{x\rightarrow7}{\lim}\frac{(x+9)-16}{{(x+9)^\frac{1}{4}}-(16)^\frac{1}{4}}$$

$$= \frac{112}{27}$$

Solution 3:

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 $x-7$s 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:

$$(4\cdot2^3)\left(\frac{1}{6}-\frac{1}{3\cdot 3^2}\right)$$