I need to evaluate $\lim_{t\to 8} \frac{t^{1/3} - 2} {t-8}$, without using L'Hôpital's Rule

Question

I'm having trouble doing this problem. The answer is $1/12$:

$$\lim_{t\to 8} \frac{t^{1/3} - 2} {t-8}$$

What I did first was: I factored out 'cube root' into:

$$\lim_{t\to 8} \left(\frac{t-8}{t^3 - 512}\right)^{1/3} $$

but kept getting the answer $1/4$ so I'm not exactly sure why that way doesn't work because it makes perfect logical sense.

Please help me without the L'Hôpital's rule.

Answer 1

You seem to be using that $$ (t-8)^3=t^3-512 \tag{*} $$ which is, unfortunately, a huge mistake. Similarly, your purported equality $$ t^{1/3}-2=(t-8)^{1/3} $$ is another mistake.

In order to see why (*) is wrong, compute both sides with $t=4$; the left hand side is $(4-8)^3=(-4)^3=-64$, whereas the right hand side is $$ (-4)^3-512=-64-512=-576 $$

The easiest way is the substitution $t^{1/3}=u$; if $t\to8$, then $u=t^{1/3}\to2$, so the limit becomes $$ \lim_{u\to2}\frac{u-2}{u^3-8}= \lim_{u\to2}\frac{u-2}{(u-2)(u^2+4u+4)}= \lim_{u\to2}\frac{1}{u^2+4u+4} $$ that has no indeterminacy. The technique is actually no different from noting that $$ t-8=(t^{1/3}-2)(t^{2/3}+4t^{1/3}+4) $$ and doing the simplification, but the algebra is perhaps more straightforward.

Answer 2

Hint

$a^3-b^3=(a-b)(a^2+ab+b^2)$

Can you see denominator now?

Set $t^{1/3}=a$ and $b=2$ You'll get : $$\frac{a-b}{a^3-b^3}=\frac{1}{a^2+ab+b^2}$$ Now put the values of $a$ and $b$

$$\lim_{t\to 8}\frac {t^{1/3}-2}{t-8} = \lim_{t\to 8}\frac {1}{({t^{1/3}})^2+(t^{1/3})2+2^2}$$

Now Indeterminacy is removed.Simply put $t=8$

$$\lim_{t\to 8}\frac {1}{({t^{1/3}})^2+(t^{1/3})2+2^2}=\frac{1}{4+4+4}=\boxed{\frac{1}{12}}$$

Answer 3

To avoid fractional exponent, Put $x=t^{\frac {1}{3} }$ and compute

$$\lim_{x\to 2}\frac {x-2}{x^3-2^3}$$

using

$$x^3-2^3=(x-2)(x^2+2x+4) .$$

After a simplification, You will get

$$\lim_{x\to2}\frac {1}{x^2+2x+4}=\frac {1}{12} $$

Answer 4

Hint:

Divide and by multiply by $(t^{2/3}+4+2t^{1/3})$

Now what can you say about $(t^{1/3}-2)(t^{2/3}+4+2t^{1/3})$?

Answer 5

I do not know why @aprojectilemotion deleted his answer, because it is, by far, the simplest solution. The limit

$$\lim_{t\to 8} \frac{t^{1/3} - 2} {t-8}$$

is, by definition, the derivative of the function $t^{1/3}$ at the point $8.$ A simple computation gives the answer.

Answer 6

HINT:

Let $t^{1/3}-2=h\implies t=(h+2)^3$ and $h\to0$

$\lim_{t\to8}\dfrac{t^{1/3}-2}{t-8}=\lim_{h\to0}\dfrac h{(h+2)^3-8}=\dfrac h{h(h^2+3h\cdot2+3\cdot2^2)}=?$

$h$ can be cancelled safely as $h\ne0$ as $h\to0$