# Evaluate limit of $\lim_{x \to 0} \frac{ (1 + x) ^{(1/5)} – (1 -x) ^ {(1/5)} } {x}$

algebra-precalculuslimitslimits-without-lhopital

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.

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 $$2.$$