# Can be solved without L'Hopital?

Can this limit be evaluated without l'hopital's rule?

$$\lim_{h\to0}\frac{\sqrt{8+h}-2}{h}$$

• sure, the 3rd square root is differentiable in 8. And your expression is (assuming $h\to 0$) just a difference quotient – Quickbeam2k1 Mar 7 '14 at 15:57
• It probably should be $h \to 0$ in the limit. – user133281 Mar 7 '14 at 15:59
• @Rogers mixedmath's solution is actually the simplest and most intuitive. Try to work it out based off the hint. – Tyler Holden Mar 7 '14 at 16:10
• l'Hopital's rule can actually be derived from other rules; it is not an "axiom." So if a limit can be solved with l'Hopital's rule, it can also be solved without l'Hopital's rule. – TMM Mar 7 '14 at 16:29
• @BenMillwood Agh! Sorry, made a typo, comment now corrected. – TheVal Mar 7 '14 at 20:36

$$x^3-y^3=(x-y)(x^2+xy+y^2)\implies x-y=\frac{x^3-y^3}{x^2+xy+y2}$$

Now we put

$$x=\sqrt{8+h}\;,\;\;y=2\implies \sqrt{8+h}-2=\frac{8+h-8}{(8+y)^{2/3}+2\sqrt{8+h}+4}\implies$$

$$\frac{\sqrt{8+h}-2}h=\frac1{(8+h)^{2/3}+2\sqrt{8+h}+4}\xrightarrow[h\to 0]{}\frac1{8^{2/3}+2\sqrt8+4}=\ldots$$

• Some net lag I just had, I didn't see your answer or Yiorgos' before I posted. No sense in having 3 identical answers. (+1) – robjohn Mar 7 '14 at 16:27

Try to let: $u= (8 + h)^{1/3}$

Thus $u^3 = 8 + h$

$h = u^3 - 8$

Then the limit becomes:

$$\lim_{u\to2}\frac{u-2}{u^3-8} = \lim_{u\to2}\frac{u-2}{(u-2)(u^2+2u+4)}=\lim_{u\to2}\frac{1}{u^2+2u+4}$$

• This seems to be the same basic idea as DonAntonio, but it is expressed much more simply and clearly. +1 – jpmc26 Mar 8 '14 at 0:34
• Haha wow, I don't go on the computer for 2 days and I see this. Thanks everyone! :) – user133458 Mar 10 '14 at 1:55
• This indeed is the same idea as in my answer but seriously simpler and nicer. I can't understand how come my answer was accepted and not this one. Anyway, +1 – DonAntonio Mar 11 '14 at 18:47

HINT:

Think about the definition of a derivative, recognize this expression as the derivative of something. (And yes, this can be found without l'Hopital's rule).

One neat way to solve some limits is to use Taylor series. If you recall the Maclaurin expansion $$(1 + x)^n = 1 + nx + \mathcal{O}(x^2)$$ you can see that $$\sqrt{8 + h} -2 = 2\sqrt{1 + h/8} -2 \approx 2\left(1 + \frac{h}{24}\right) -2 = \frac{h}{12}$$ so the limit becomes $$\lim_{h \to 0} \frac{h/12}{h} = \frac{1}{12}.$$

• It's a neat way to solve a lot of limits, I have never touched L'Hopital & co. since I knew Taylor expansions and small-oh notation. – Matteo Italia Mar 8 '14 at 1:28
• @MatteoItalia it's the same for me. – giordano Mar 8 '14 at 10:06

\begin{align} \frac{\sqrt{8+h}-2}{h}&= \frac{\sqrt{(8+h)^2}+2\sqrt{8+h}+4}{\sqrt{(8+h)^2}+2\sqrt{8+h}+4} \cdot \frac{\sqrt{8+h}-2}{h} \\&=\frac{(8+h)-8}{h} \cdot\frac{1}{\sqrt{(8+h)^2}+2\sqrt{8+h}+4} \to\frac{1}{12}. \end{align}