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let $f(x)=x^3$

How do I solve this limit? $$\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$$

I can replace the function with its content

$$\lim_{h\to0}\dfrac{(x+h)^3-x^3}{h}$$

Then expand the paranthesis $$\lim_{h\to0}\dfrac{x^3+3h^2x+3hx^2+h^3-x^3}{h}$$

Thus simplifying to $$\lim_{h\to0}\dfrac{3h^2x+3hx^2+h^3}{h}$$

From here, I don't know where to go in order to solve this limit. I feel like I made the problem more complex than actually simplifying it.

Cedric Martens
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2 Answers2

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You are almost done indeed note that form here

$$\lim_{h\to0}\dfrac{3h^2x+3hx^2+h^3}{h}=\lim_{h\to0} \,(3hx+3x^2+h^2)=0+3x^2+0=3x^2$$

that is exactly the derivative for $x^3$.

user
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  • I see how you passed from $$\lim_{h\to0}\dfrac{3h^2x+3hx^2+h^3}{h}$$ to $$\lim_{h\to0} ,(3hx+3x^2+h^2) $$ But I don't understand how you got to $$3x^2$$ – Cedric Martens Jun 11 '18 at 21:06
  • @CedricMartens simply substitute $h=0$ since the function is continuous at $h=0$. – user Jun 11 '18 at 21:07
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    @CedricMartens: how much is $3\cdot0\cdot x+3x^2+0^2$ ? –  Jun 11 '18 at 21:07
  • @CedricMartens (By the way, if you fix $x$ you're viewing $3hx + 3x^2 + h^2$ as a polynomial in $h$, i.e. $ah^2 + bh + c$. Polynomials in general are continuous - you can prove this inductively - so $\lim_{h \to 0} p(h) = p(0)$. So, like gimusi says, you just substitute with $h = 0$. Since the claim is true no matter which $x$ you fix, it's true for all $x$.) – Chris Jun 11 '18 at 21:28
  • When we take the limit as $h\to 0$ we are allowing $h$ to become arbitrarily small. So small, in fact, that it is effectively 0. – Doug M Jun 11 '18 at 22:10
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The limit is not in two variables, because $x$ is constant. When you arrive at $$ \lim_{h\to0}\frac{3hx^2+3h^2x+h^3}{h}=\lim_{h\to0}(3x^2+3xh+h^2) $$ you can apply standard rules of limits: \begin{align} \lim_{h\to0}(3x^2+3xh+h^2) &=\lim_{h\to0}3x^2+\lim_{h\to0}3xh+\lim_{h\to0}h^2\\[6px] &=3x^2+3x\lim_{h\to0}h+\Bigl(\,\lim_{h\to0}h\Bigr)^{\!2}\\[6px] \end{align} provided all limits after the top equals sign exist finite. Since they do, you can conclude with $$ =3x^2 $$ Usually this is applied without doing the splitting into “more elementary” limits, because the same rules applied before tell us that polynomials are continuous functions, so their limit at a point can be computed by simple substitution.

egreg
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