# Evaluating $\lim_{h \to 0}\frac{(x+h)^{\frac15}-x^{\frac15}}{h}$

The limit is: $$\lim_{h \to 0}\frac{(x+h)^{\frac15}-x^{\frac15}}{h}$$

When I use calculator and substitute $h$ with $0.000001$ and $-0.000001$, the result is:

$$\frac{1}{5x^{\frac45}}$$

My question is:

1. How to do it without calculator.

2. Show me the steps on how it's being done.

– Vim
Feb 7 '15 at 6:22
• @Vim doing it with calculator is very simple. Input those numbers to a scientific dot matrix calculator and hit '='. Substitute h with 0.000001 and note the answer, next substitute h with -0.000001 and note the answer. Finally compare the two answers Feb 7 '15 at 6:25
• But unless you have presumed that the result is something like $ax^b$ how could you reach the exact coefficient $\frac15$ and exponent $-\frac45$ only by these trials? I don't think it's very easy to do
– Vim
Feb 7 '15 at 6:29
• @Vim OP would never pass the tortoise with his approach ;) Feb 7 '15 at 6:34
• @induktio I mean, in fact, first we gotta know where the tortoise is before we are about to approach...
– Vim
Feb 7 '15 at 6:39

Remember that $$a^5 - b^5 = (a-b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4).$$ Then, $$a-b = \frac{a^5-b^5}{a^4 + a^3b + a^2b^2 + ab^3 + b^4}$$ Taking $a = (x+h)^{1/5}$ and $b=x^{1/5}$ we get that \begin{align} \lim_{h \to 0}\frac{(x+h)^{1/5}-x^{1/5}}{h} &= \lim_{h \to 0}\frac{(x+h) -x}{h(a^4 + a^3b + a^2b^2 + ab^3 + b^4)} \\ &= \lim_{h \to 0}\frac{1}{(a^4 + a^3b + a^2b^2 + ab^3 + b^4)} \\ &= \frac{1}{5x^{4/5}}, \end{align} since $\lim_{h\to 0}a = \lim_{h\to 0}b = x^{1/5}$.

• Thank you. You managed to solve this problem without derivatives which I haven't fully understand yet. Feb 7 '15 at 13:52
• @Kristian It's quite obvious that if you were familiar with the concept you wouldn't be asking the question :) Feb 7 '15 at 21:46
• @hjhjhj57 I think the last line should end with $x^{1/5}$, not $x$. Feb 8 '15 at 1:41
• @Théophile edited. Feb 8 '15 at 10:33
• @hjhjhj57 This is one of the rare high quality answers, good job! Feb 8 '15 at 10:42

Recall how the derivative of a function $f(x)$ is defined: $$f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}.\tag{1}$$ In the context of your problem, it is fairly clear that we are dealing with $(1)$ where $f(x)=x^{1/5}$, giving us $$f'(x) = \lim_{h\to 0}\frac{(x+h)^{1/5}-x^{1/5}}{h}.\tag{2}$$ I assume you have encountered the Power Rule in calculus. Using this, we can see how to calculate $(2)$: $$\lim_{h\to 0}\frac{(x+h)^{1/5}-x^{1/5}}{h} = \frac{d}{dx}x^{1/5} = \frac{1}{5}x^{-4/5}=\frac{1}{5x^{4/5}},$$ as desired.

The expression you have given is the definition for derivative of $x^{1/5}$ with respect to $x$. $$\lim_{h\to 0}\frac{(x+h)^{1/5}-x^{1/5}}{h}=\frac{d}{dx}x^{1/5}=\frac{1}{5}x^{1/5-1}=\frac{1}{5}x^{-4/5}=\frac{1}{5x^{4/5}}$$

I guess you don't know (yet) about derivatives. Of course the answer to your question depends a bit on what you know already. Here is an answer relying on the generalized binomial theorem $$(a+b)^y = \sum_{i=0}^\infty \binom{y}{i} a^i b^{y-i} \tag{1}$$ with $$\binom{y}{i}= \frac{y(y-1)\cdots(y-i+1)}{i!}.$$

In your case, you have $a=h$, $b=x$, and $y=1/5$ which yields $$(h+x)^{1/5}-x^{1/5} = \sum_{i=1}^\infty \binom{1/5}{i} h^i x^{1/5-i}.$$ As a result, you can write $$\frac{(h+x)^{1/5}-x^{1/5}}{h} = \sum_{i=1}^\infty \binom{1/5}{i} h^{i-1} x^{1/5-i}.$$

Now under the limit $h\to 0$ only the term with $i=1$ survives and you simply have to evaluate $$\lim_{h\to0}\frac{(h+x)^{1/5}-x^{1/5}}{h} = \binom{1/5}{1} x^{1/5-1} = \frac15 x^{-4/5},$$

use the binomial theorem in the form $$(BIG + small)^n = BIG^n + nBIG^{n-1}small+ \cdots$$

here is how it is applied: $(x + h)^{1/5} = x^{1/5} + \frac{1}{5}x^{4/5}h + \cdots$ so that $\dfrac{(x+h)^{1/5} - x^{1/5}}{h} = \frac{1}{5}x^{4/5}+ \cdots$ and in the limit you $\frac{1}{5}x^{4/5}$

$$\lim_{h \to 0} \frac{(x(1+\frac{h}{x}))^{1/5} -x^{1/5}}{h} = \lim_{h \to 0} \frac{x^{1/5}(1+\frac{h}{x})^{1/5} -x^{1/5}}{h} = \lim_{h \to 0} \bigl( x^{1/5} \cdot \frac{(1+\frac{h}{x})^{1/5} -1}{h} \bigr) = x^{1/5}\lim_{h \to 0} \frac{(1+\frac{h}{x})^{1/5} -1}{h} =x^{1/5}\lim_{h \to 0} \frac{e^{1/5 \cdot\ln{(1+\frac{h}{x})}} -1}{h}$$

Since $\lim\limits_{h \to 0} \frac{1}{5} \cdot \ln{(1+\frac{h}{x})}=0$,

$$x^{1/5}\lim_{h \to 0} \frac{e^{1/5 \cdot\ln{(1+\frac{h}{x})}} -1}{h} = x^{1/5}\lim_{h \to 0} \frac{e^{1/5 \cdot\ln{(1+\frac{h}{x})}} -1}{1/5 \cdot\ln{(1+\frac{h}{x})}} \cdot \frac{1/5 \cdot\ln{(1+\frac{h}{x})}}{h} = \\ x^{1/5}\lim_{h \to 0} \frac{e^{1/5 \cdot\ln{(1+\frac{h}{x})}} -1}{1/5 \cdot\ln{(1+\frac{h}{x})}} \cdot \lim_{h \to 0} \frac{1/5 \cdot\ln{(1+\frac{h}{x})}}{\frac{h}{x}} \cdot \lim_{h \to 0} \frac{1}{x} = x^{1/5} \cdot 1 \cdot 1/5 \cdot \frac{1}{x} = \frac{x^{1/5-1}}{5}$$