How do you calculate lims that are 0/0 without L'Hospital's law?

Question

*I prefer hint more than a complete solution if possible.

edit: Let the second equation be x, can I do $1/x$ and solve with polynomial division?

Hi,

for those two lims, we get lim (0/0) and we are not allowed to use L'Hospital's law in order to solve.

1.$ \lim_{x\rightarrow10}\frac{x-10}{\sqrt{3x+6}-6}$

2.$\lim_{x\rightarrow(-2)}\frac{4x^3-15x+2}{x^4-3x-22}$

Do you have a hint for me?

Thanks

Answer 1

Factoring and conjugation are two methods that you can use to eliminate the removable discontinuities.

1. $\lim\limits_{x\to 10} \dfrac{x-10}{\sqrt{3x+6}-6}\cdot\dfrac{\sqrt{3x+6}+6}{\sqrt{3x+6}+6}$

2. $\lim\limits_{x\to -2}\dfrac{4x^3-15x+2}{x^4-3x-22}=\dfrac{(x+2)(\cdots)}{(x+2)(\cdots)}$

Answer 2

1) Rationalize the denominator.

2) Since the numerator and denominator are both $0$ when $x=-2$, you can conclude that $(x+2)$ is a factor of both by the Remainder Theorem. Factor and cancel out that term.

Answer 3

For $\lim_{x\rightarrow(-2)}\frac{4x^3-15x+2}{x^4-3x-22} $, let $x = y-2$.

This becomes

$\begin{array}\\ \lim_{y\to 0}\frac{4(y-2)^3-15(y-2)+2}{(y-2)^4-3(y-2)-22} &=\lim_{y\to 0}\frac{y (4 y^2 - 24 y + 33)}{y (y^3 - 8 y^2 + 24 y - 35)}\\ &=\lim_{y\to 0}\frac{4 y^2 - 24 y + 33}{ y^3 - 8 y^2 + 24 y - 35}\\ &=-\frac{33}{ 35}\\ \end{array} $

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(added later)

And for $\lim_{x\rightarrow10}\frac{x-10}{\sqrt{3x+6}-6} $, let $y = x-10$. This becomes

$\begin{array}\\ \lim_{y\rightarrow 0}\frac{y}{\sqrt{3(y+10)+6}-6} &=\lim_{y\rightarrow 0}\frac{y}{\sqrt{3y+36}-6}\\ &=\lim_{y\rightarrow 0}\frac{y}{6\sqrt{1+y/12}-6}\\ &=\lim_{y\rightarrow 0}\frac{y}{6(1+y/24+O(1/y^2))-6}\\ &=\lim_{y\rightarrow 0}\frac{y}{y/4+O(1/y)}\\ &=4\\ &\text{or}\\ &=\lim_{y\rightarrow 0}\frac{y}{\sqrt{3y+36}-6}\frac{\sqrt{3y+36}+6}{\sqrt{3y+36}+6}\\ &=\lim_{y\rightarrow 0}\frac{y(\sqrt{3y+36}+6)}{(3y+36)-36}\\ &=\frac{12}{3}\\ &=4\\ \end{array} $

Answer 4

Also, by the definition of a derivative we obtain: $$\lim_{x\rightarrow10}\frac{x-10}{\sqrt{3x+6}-6}=\frac{1}{\sqrt3}\cdot\frac{1}{\lim\limits_{x\rightarrow10}\frac{\sqrt{x+2}-\sqrt{12}}{x-10}}=\frac{1}{\sqrt3}\cdot\frac{1}{\left(\sqrt{x+2}\right)'_{x=10}}=\frac{1}{\sqrt3}\cdot2\sqrt{10+2}=4.$$