0
$\begingroup$

I got this question asked in school MCQ and couldn't answer it. Here's the question $$\lim_{x \to 0} \frac{ (1 + x) ^{(1/5)} - (1 -x) ^ {(1/5)} } {x} $$

There were 4 options a)1/5 b)2/5 c)-1/3 d)0

I thought about applying $$\lim_{x \to a} \frac {x ^ n - a ^ n } {x -a} = n a^{(n-1)}$$ but x tends to 0 not (1-x). Also, $ (1 + x) - 1 = x$ but $ (1 + x) - (1 - x) = 2x $ I cant think any other way to solve it.

$\endgroup$
3
  • $\begingroup$ Use the Binomial Theorem. $\endgroup$ Commented Nov 9, 2021 at 14:38
  • $\begingroup$ Just to be clear, the competition is over now, right? $\endgroup$ Commented Nov 9, 2021 at 14:38
  • $\begingroup$ Competition is now over, also binomial theorem is 😬 $\endgroup$
    – Machinexa
    Commented Nov 9, 2021 at 14:39

2 Answers 2

4
$\begingroup$

Start by adding and subtracting $1$ in the numerator:

$$\lim_{x \to 0} \frac{(1 + x)^{\frac15} - 1 + 1 - (1 - x)^{\frac15}}{x}$$

Now split up the limit like this:

$$\lim_{x \to 0} \frac{(1 + x)^{\frac15} - 1}x + \lim_{x \to 0}\frac{(1 -x)^{\frac15} - 1}{-x}$$

Using the substitution $u = -x$ on the second limit, you should see that both limits are the definition of the derivative of $f(x) = x^{\frac15}$ at $x = 1$:

$$f'(x_0) = \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h}$$ $$f'(1) = \lim_{h \to 0} \frac{(1 + h)^{\frac15} - 1^{\frac15}}{h}$$

Using the power rule, we know that $f'(x) = \frac15x^{-\frac45}.$ So, the answer is $2f'(1) = 2(\frac15(1)^{-\frac45}) = \boxed{\frac25.}$

Alternatively, you can see this by using the alternative definition $f'(1) = \lim_{x \to 0}\frac{f(1 + x) - f(1 - x)}{2x}$ and multiplying the numerator and denominator of our original limit by $2.$

$\endgroup$
6
  • $\begingroup$ I understood how you separated those two limits keeping minus at bottom, but didnt get step after it. Would you share some links to read before reading this answer as i am very beginner especially the 2f' and the derivative part. I think i dont understand core of derivative but know how to solve it $\endgroup$
    – Machinexa
    Commented Nov 9, 2021 at 14:48
  • $\begingroup$ @Machinexa I added some clarifications, does that help clear things up? $\endgroup$ Commented Nov 9, 2021 at 14:51
  • $\begingroup$ Yes, now it seems lot more easy $\endgroup$
    – Machinexa
    Commented Nov 9, 2021 at 14:52
  • 1
    $\begingroup$ Awesome, glad to hear it. How did the competition go by the way? $\endgroup$ Commented Nov 9, 2021 at 15:01
  • 1
    $\begingroup$ Interesting, for some reason I had it in my head that MCQ stood for MathCounts Qualifiers. (not actually sure if that's an abbreviation they use, I've never actually been) In any case, great job, that's a solid placement $\endgroup$ Commented Nov 9, 2021 at 15:05
1
$\begingroup$

Since this was a multiple choice question, there are no restrictions regarding the methods you use to get the answer. So, you can use L'Hôpital's rule and differentiate both the numerator and denominator ($\frac 00$ indetermination): $$ \lim_{x\to 0}\dfrac{(1+x)^{1/5}-(1-x)^{1/5}}{x} = \lim_{x\to 0}\dfrac{\frac 15(1+x)^{-4/5}-(-\frac 15 )(1-x)^{-4/5}}{1} = \frac{1/5+1/5}{1} = \frac 25. $$


$$\left((1+x)^{1/5}\right)' = \frac 15 (1+x)' (1+x)^{1-\frac 15} = \frac 15 (1+x^{-4/5})$$

$$\left((1-x)^{1/5}\right)' = \frac 15 (1-x)' (1-x)^{1-\frac 15} = \frac 15 (-1)(1+x^{-4/5})$$

Please note that in many courses you are not allowed to use this rule, but rather perform a reduction to notable limits or the identification with the definition of a derivative (as did @StephenDonovan).

$\endgroup$
1
  • 1
    $\begingroup$ Also a good method, and definitely a method worth knowing. For OP, since you mentioned that the differentiation is a bit new to you, I think it's worth clarifying that the extra negative on the second term in the numerator comes from the chain rule $\endgroup$ Commented Nov 9, 2021 at 15:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .