Evaluating $\lim_{x \to 7} \frac{\sqrt{x-3}-2}{\sqrt{x+2}-10+x}$without L'Hopital's rule

Question

Been trying to solve this one on several occasions and can't manage it, my prof disallows the use of L'Hopital's, can someone please explain the trick here? thank you in advance! $$\lim_{x \to 7} \frac{\sqrt{x-3}-2}{\sqrt{x+2}-10+x}$$

this is in the context of using various tricks to get rid of the $0/0$ paradox, the other tasks were stuff along the lines of taking common factors and getting rid of them or doing that after multiplying the numerator and the denominator by a value that would make the $(a-b)(a+b)=a^2-b^2$ formula to get rid of square roots, which I tried, multiplying them by $(\sqrt{x-3}+2)(\sqrt{x+2}+10+x)$, but that didn't seem to go anywhere. We were also introduced to some cases of wonderful limits (or however that's called in English) but, since $x$ is approaching $7$, I assume it's of no use?

Answer 1

I wrote this out step by step to make everything clear, but most of this could have been written in a few lines.

For all $x \ge 3$ and $x \ne 7$, $$\begin{align} \frac{\sqrt{x-3} - 2}{\sqrt{x+2}-10+x} &= \frac{\sqrt{x-3} - 2}{\sqrt{x+2}-(10-x)} \\ \\ &= \frac{(\sqrt{x-3} - 2)(\sqrt{x+2} + (10-x))}{(\sqrt{x+2}-10+x)(\sqrt{x+2} + (10-x))} \\ \\ &= \frac{(\sqrt{x-3} - 2)(\sqrt{x+2} + (10-x))}{x+2-(10+x)^2} \\ \\ &= \frac{(\sqrt{x-3} - 2)(\sqrt{x+2} + (10-x))}{(x-7)(14-x)} \\ \\ &= \frac{(\sqrt{x-3} - 2)(\sqrt{x-3} + 2)(\sqrt{x+2} + (10-x))}{(x-7)(14-x)(\sqrt{x-3} + 2)} \\ \\ &= \frac{((x-3) - 2^2)(\sqrt{x+2} + (10-x))}{(x-7)(14-x)(\sqrt{x-3} + 2)} \\ \\ &= \frac{(x-7)(\sqrt{x+2} + (10-x))}{(x-7)(14-x)(\sqrt{x-3} + 2)} \\ \\ &= \frac{\sqrt{x+2} + 10-x}{(14-x)(\sqrt{x-3} + 2)}. \end{align}$$

Now take the limit as $x \to 7$.

Answer 2

Making use of $(a+b)(a-b)=a^2-b^2$ does work here. Informally speaking, the fact that $7$ is a root of both the numerator and the denominator tells us that $(x-7)$ is kind of a factor in both of them, so we just need to extract it from both to be able to cancel it out.

In your attempt, it appears that you got some of your signs wrong, though. You should see the denominator as $\sqrt{x+2}-(10-x)$, and then this trick is going to work:

$$\lim_{x\to7}\frac{\sqrt{x-3}-2}{\sqrt{x+2}-(10-x)} = \lim_{x\to7}\frac{(\sqrt{x-3}-2)(\sqrt{x-3}+2)(\sqrt{x+2}+(10-x))}{(\sqrt{x+2}-(10-x))(\sqrt{x-3}+2)(\sqrt{x+2}+(10-x))} = \ldots,$$

and you should be able to take it from here.

Answer 3

Here is a slightly nonstandard method but a short one and with no square roots on the way. Let $s=\sqrt{x-3},$ $t=\sqrt{x+2}.$ Then $s^2-4=t^2-9.$ As $x=t^2-2$ the expression takes the form $${s-2\over t-10+t^2-2}={s-2\over t-3}{1\over t+4}\\ ={s^2-4\over t^2-9}{t+3\over s+2}{1\over t+4} ={t+3\over (s+2)(t+4)}\underset{x\to 7}{\longrightarrow} {6\over 4\cdot 7 }={3\over 14}$$

Answer 4

An idea: first a substitution to make things a little simpler:

$$t=x-7\implies \lim_{x\to 7}\frac{\sqrt{x-3}-2}{\sqrt{x+2}-(10-x)}=\lim_{t\to0}\frac{\sqrt{t+4}-2}{\sqrt{t+9}+(t-3)}$$

And now some basic albeit messy algebra:

$$\frac{\sqrt{t+4}-2}{\sqrt{t+9}+(t-3)}\cdot\frac{\sqrt{t+4}+2}{\sqrt{t+4}+2}=\frac t{\left(\sqrt{t+9}+(t-3)\right)\left(\sqrt{t+4}+2\right)}=$$

$$=\frac t{\left(\sqrt{t+9}+(t-3)\right)\left(\sqrt{t+4}+2\right)}\cdot\frac{\sqrt{t+9}-(t-3)}{\sqrt{t+9}-(t-3)}=$$

$$\frac{t\left(\sqrt{t+9}-(t-3)\right)}{\left(t+9-(t-3)^2\right)\left(\sqrt{t+4}+2\right)}=\frac{t\left(\sqrt{t+9}-(t-3)\right)}{\left(-t(t-7)\right)\left(\sqrt{t+4}+2\right)}=$$

$$=-\frac{\sqrt{t+9}-(t-3)}{\left((t-7)\right)\left(\sqrt{t+4}+2\right)}\xrightarrow[t\to0]{}-\frac{6}{(-7)\cdot4}=\frac3{14}$$

Answer 5

One could observe that $$\begin{align} \sqrt{x+2}-10+x &=(x+2)+\sqrt{x+2}-12 \\ &=y^2+y-12 \qquad y:=\sqrt{x+2}\\ &=(y+4)(y-3) \end{align}$$ So, the expression can be written $$\frac{1}{\sqrt{x+2}+4}\cdot\frac{\sqrt{x-3}-2}{\sqrt{x+2}-3}$$ The first factor isn't a problem, but the second gives $0/0$ when $x=7$. Multiplying the numerator and denominator of that second factor by $(\sqrt{x-3}+2)(\sqrt{x+2}+3)$ will provide a convenient cancellation.

This is conceptually the same as what OP attempted, but involves less-complicated factors.