$$\lim_\limits{x\to-1}\frac{\sqrt[3]{26-x}-\sqrt{8-x}}{3x^2+4x+1}$$
I was asked to solve this problem without using l'hopital's rule. Using Mathematica it shows that we should rationalize first but the algebra gets quite nasty near the end so I was wondering if there was a better way of approaching this problem...
On
Here is just a short sketch of a not so cumbersome algebraic solution.
Looking at $26=3^3-1$ and $8=3^2-1$, intuition may guide you to set $x=y-1$ and consider
$$\lim_\limits{x\to-1}\frac{\sqrt[3]{26-x}-\sqrt{8-x}}{3x^2+4x+1} = \lim_\limits{y\to \color{blue}{0}}\underbrace{\frac{\sqrt[3]{27-y}-\sqrt{9-y}}{3(y-1)^2+4(y-1)+1}}_{=\frac{\sqrt[3]{27-y}-\sqrt{9-y}}{(3y-2)y}}$$
This looks much nicer and, since $\lim_{y\to 0}\frac 1{3y-2}=-\frac 12$, we only need to consider
$$\frac{\sqrt[3]{27-y}-\sqrt{9-y}}{y}= \underbrace{\frac{\sqrt[3]{27-y}-3}{y}}_{= -\frac 1{\sqrt[3]{(27-y)^2} + \sqrt[3]{(27-y)\cdot 27} +\sqrt[3]{27^2}}} - \frac{\sqrt{9-y}-3}{y}$$
The last term you may handle with the standard binomial formula.
So, putting all together you will get (not forgetting $\frac 1{3y-2}$)
$$-\frac 12\left(-\frac 1{27} + \frac 16\right) = -\frac 7{108}$$
On
Using $$\lim\limits_{x \to 0}\frac{\sqrt[m]{1+x}-1}{x}=\frac{1}{m}$$ we can write $$\lim_\limits{x\to-1}\frac{\sqrt[3]{26-x}-\sqrt{8-x}}{3x^2+4x+1}=\lim_\limits{x\to-1}\frac{\sqrt[3]{27-(x+1)}-\sqrt{9-(x+1)}}{(x+1)(3x+1)}=\\ =3\lim_\limits{x\to-1}\frac{\left(\sqrt[3]{1+\frac{x+1}{-27}}-1\right)-\left(\sqrt{1+\frac{x+1}{-9}}-1 \right)}{(x+1)(3x+1)}=\\ =3\lim_\limits{x\to-1}\frac{\frac{\left(\sqrt[3]{1+\frac{x+1}{-27}}-1\right)}{\frac{x+1}{-27}}\cdot\frac{x+1}{-27}- \frac{\left(\sqrt{1+\frac{x+1}{-9}}-1 \right)}{\frac{x+1}{-9} }\cdot\frac{x+1}{-9}}{(x+1)(3x+1)}= 3\frac{\frac{1}{3}\cdot \frac{1}{-27}-\frac{1}{2}\cdot \frac{1}{-9}}{-2}=\\ =-\frac 12\left(-\frac 1{27} + \frac 16\right) $$
On
Let $f(z)=(1+z)^{2/3}+(1+z)^{1/3}+1.$ We have $[(1+z)^{1/3}-1]f(z)=z.$ So $(1+z)^{1/3}=1+z/f(z).$ We have $1/f(z)\to 1/3$ as $z\to 0$ so let $1/f(z)=(1 +g(z))/3 $ where $g(z)\to 0$ as $z\to 0.$ Then $$(1+z)^{1/3}=1+(z/3)(1+g(z)).$$
For $|z|<1$ let $h(z)=(1+z)^{1/2}+1.$ We have $[(1+z)^{1/2}-1]h(z)=z.$ So $(1+z)^{1/2}=1+z/h(z).$ We have $1/h(z)\to 1/2$ as $z\to 0$ so let $1/h(z)=(1+j(z))/2 $ where $j(z)\to 0$ as $z\to 0.$ Then $$(1+z)^{1/2}=1+(z/2)(1+j(z)).$$
Let $x=-1+y.$
For $|y|$ small enough that $(1+y/9)^{1/2}\in \Bbb R$ we have $$(26-x)^{1/3}-(8-x)^{1/2}=(27+y)^{1/3}-(9+y)^{1/2}=$$ $$=3 (1+y/27)^{1/3}-3(1+y/9)^{1/2}=$$ $$=3[1+(y/81)(1+g(y/27))]-3[1+(y/18)(1+j(y/9))]=$$ $$= 3(1+y/81)-3(1+y/18)+yk(y)=$$ $$=y(-7/54+k(y))$$ where $k(y)\to 0$ as $y\to 0,$ i.e. as $x\to -1.$
FYI. $k(y)=g(y/27)/27-j(y/9)/6.$
The denominator in the Q is $3x^2+4x+1=(x+1)(3x+1)=y(3y-2).$ So the whole expression is $\frac {y(-7/54+k(y))}{y(3y-2)}=\frac {-7/54+k(y)}{3y-2}$ with $y\to 0$ and $k(y)\to 0.$
$\frac {(26-x)^\frac 13 - (8-x)^\frac 12}{(3x + 1)(x+1)}$
We are going to multiply top and bottom by a factor that will clear out the fractional roots.
Note that:
$a^6 - b^6 = (a-b)(a^5+a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)$
Let $a = (26-x)^\frac 13$ Let $b = (8-x)^\frac 12$
$\frac {(a-b)(a^5+a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)}{(3x+1)(x+1)(a^5+a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)}\\ \frac {a^6-b^6}{(3x+1)(x+1)(a^5+a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)}\\ \frac {(26-x)^2 - (8-x)^3}{(3x+1)(x+1)(a^5+a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)}$
Lets multiply out and simplify the numerator. We will ignore the denominator for now.
$(26-x)^2 - (8-x)^3\\ 676 - 52x + x^2 - (512 - 192x + 24x^2 - x^3)\\ x^3 - 23x^2 +140x + 164$
We knew from the beginning that the numerator equaled zero when $x - 1$ Multiplying by our factor to rationalize the numerator doesn't change that. $(x+1)$ is a factor of the numerator.
$x^3 - 23x^2 +140x + 164 = (x+1)(x^2 - 24x + 164)$
$\lim_\limits{x\to-1}\frac {(x+1)(x^2 - 24x + 164)}{(3x+1)(x+1)(a^5+a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5)}$
We can cancel the $x+1$ factor, and evaluate the rest at $x = -1.$ When $x = 1$ both $a$ and $b$ equal $3.$
$\frac {189}{(-2)(3^5)(6)} = -\frac {7}{108}$