I'm trying to solve the following limit.
Unfortunately, I haven't made any headway.
I tried multiplying by the conjugates of both the numerator and denominator (in 2 different attempts, of course), but ended up with a monster of an expression either time.
On
You can apply L'Hospital Rule because after putting limit, we get, $$\frac{0}{0}$$ Applying L'Hospital Rule, $$\lim_{x\to2}\frac{\frac{-1}{2\sqrt{6-x}}}{\frac{-1}{2\sqrt{3-x}}}$$ After simplifying, you get, $$\lim_{x\to2}\frac{\sqrt{3-x}}{\sqrt{6-x}}$$ Applying limits, $$\frac{1}{2}$$
On
HINT:
$$\lim_{x\to2}\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\lim_{x\to2}\dfrac{6-x-2^2}{3-x-1^2}\cdot\lim_{x\to2}\dfrac{\sqrt{3-x}+1}{\sqrt{6-x}+2}$$
On
I like to let my variables approach zero. So I would do this like this:
$\begin{array}\\ \lim \limits_{x \to 2}\dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1} &=\lim \limits_{y \to 0}\dfrac{\sqrt{6-(y+2)}-2}{\sqrt{3-(y+2)}-1} \qquad(x = y+2)\\ &=\lim \limits_{y \to 0}\dfrac{\sqrt{4-y}-2}{\sqrt{1-y}-1}\\ &=\lim \limits_{y \to 0}\dfrac{2\sqrt{1-y/4}-2}{\sqrt{1-y}-1}\\ &=\lim \limits_{y \to 0}\dfrac{2(1-y/8+O(y^2))-2}{(1-y/2+O(y^2))-1}\\ &=\lim \limits_{y \to 0}\dfrac{-y/4+O(y^2)}{-y/2+O(y^2)}\\ &=\lim \limits_{y \to 0}\dfrac{-1/4+O(y)}{-1/2+O(y)}\\ &=\dfrac12\\ \end{array} $
Let $y = 3 - x$. Then we have \begin{align} \lim_{x \to 2} \frac{\sqrt{6-x} - 2}{\sqrt{3 - x} - 1} &= \lim_{y \to 1} \frac{\sqrt{y + 3} - 2}{\sqrt{y} - 1}\\ &= \lim_{y \to 1} \frac{(y + 3 - 2^2)(\sqrt{y} + 1)}{(y - 1)(\sqrt{y + 3} + 2)}\\ &= \lim_{y \to 1} \frac{(y - 1)(\sqrt{y} + 1)}{(y-1)(\sqrt{y + 3} + 2)}\\ &= \lim_{y \to 1} \frac{\sqrt{y} + 1}{\sqrt{y + 3} + 2} = \frac{1 + 1}{2 + 2} = \frac{1}{2} \end{align}