Can this question be solved in any way other than using l'hopital's rule?

Question

$$\lim\limits_{x \to -2} \frac{x+2}{(x-30)^{1/5} - (x^2 + 12)^{1/4} + 4}$$

Answer 1

It certainly can be.

Write $f(x) = (x - 30)^{1/5} - (x^2 + 12)^{1/4}$. Note that $f(-2) = -4$.

Note that $\lim\limits_{x \to -2} \frac{(x - 30)^{1/5} - (x^2 + 12)^{1/4} + 4}{x + 2} = \lim\limits_{x \to -2} \frac{f(x) - f(-2)}{x - -2}$ is, by definition, $f'(-2)$.

Therefore, the above limit is equal to $f'(-2)$. Now $f'(x) = \frac{1}{5} (x - 30)^{-4/5} - 2x\frac{1}{4}(x^2 + 12)^{-3/4}$, so we see that $f'(-2) = \frac{1}{80} + \frac{1}{8} = \frac{11}{80}$.

Therefore, $\lim\limits_{x \to -2} \frac{x + 2}{(x - 30)^{1/5} - (x^2 + 12)^{1/4} + 4} = \frac{1}{\lim\limits_{x \to -2} \frac{(x - 30)^{1/5} - (x^2 + 12)^{1/4} + 4}{x + 2}} = \frac{1}{\frac{11}{80}} = \frac{80}{11}$.

Answer 2

Want $\lim\limits_{x \to -2} f(x) $ where $f(x) =\frac{x+2}{(x-30)^{1/5} - (x^2 + 12)^{1/4} + 4} $.

Let $x = y-2$.

We just need the generalized binomial theorem.

$\begin{array}\\ f(x) &=f(y-2)\\ &=\frac{x+2}{(x-30)^{1/5} - (x^2 + 12)^{1/4} + 4}\\ &=\frac{y}{(y-32)^{1/5} - ((y-2)^2 + 12)^{1/4} + 4}\\ &=\frac{y}{(y-32)^{1/5} - (y^2-4y+16)^{1/4} + 4}\\ &=\frac{y}{(-2)(1-(y/32))^{1/5} - 2(1-(y/4)+(y^2/16))^{1/4} + 4}\\ &=\frac{y}{(-2)(1-(y/(160))+O(y^2)) - 2(1-(y/16)+O(y^2)) + 4}\\ &=\frac{y}{(-2)+(y/(80))+O(y^2) - 2+(y/8)+O(y^2) + 4}\\ &=\frac{y}{y/(80)+(y/8)+O(y^2)}\\ &=\frac{y}{11y/(80)+O(y^2)}\\ &=\frac{1}{11/(80)+O(y)}\\ &=\dfrac{80}{11}+O(y)\\ &\to\dfrac{80}{11} \qquad\text{as }y \to 0\\\\ \end{array} $

Answer 3

Let us write $x=-2+h$. We use the formulas: $$ \begin{aligned} A^5-B^5&=(A-B)(A^4+A^3B+A^2B^2 +AB^3+B^4)\ ,\\ A^4-B^4&=(A-B)(A^3+A^2B+AB^2+B^3) \end{aligned} $$ below. We have: $$ \begin{aligned} \frac{2+(x-30)^{1/5}}{x+2} &= \frac{32^{1/5}-(32-h)^{1/5}}h \\ &= \frac{32^{1/5}-(32-h)^{1/5}}{32-(32-h)} \\ &=\frac1{32^{4/5}+32^{3/5}(32-h)^{1/5}+\dots+(32-h)^{1/5}} \\ &\to\frac 1{5\cdot 16}=\frac 1{80}\qquad\text{ for }h\to0\ , \\[3mm] \frac{2-(x^2+12)^{1/4}}{x+2} &= \frac{16^{1/4}-(h^2-4h+16)^{1/4}}h \\ &= \frac {16^{1/4}-(h^2-4h+16)^{1/4}} {16-(h^2-4h+16)} \cdot \frac{16-(h^2-4h+16)} h \\ &= \frac 1{16^{3/4} +\dots+(h^2+4h+16)^{3/4}}\cdot (4-h) \\ &\to \frac 1{4\cdot 8}\cdot 4=\frac 18 \qquad\text{ for }h\to0\ , \end{aligned} $$ so the sum of the two expressions $$ \frac{2+(x-30)^{1/5}}{x+2} + \frac{2-(x^2+12)^{1/4}}{x+2} $$ converges to the sum $$ \frac1{80}+\frac 18=\frac{1+10}{80}=\frac {11}{80}\ . $$ It remains to take the reversed fraction...