# Strange limit problem to be solved without Hospital's Rule…?

Having trouble solving this limit problem without L'Hôpital's Rule...

$$\lim_{x \to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$

Tried multiplying the function by the conjugate/inverse-conjugate, of both the numerator and denominator... but no avail.... any ideas?

• why not use L'Hospital's? The reason it exists is because there are limits you can't solve without it. – setholopolus Dec 15 '16 at 19:41
• You're right you should multiply by the conjugate,where is it that you have a problem? – kingW3 Dec 15 '16 at 19:42

$$\lim_{x \to 2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\lim_{x \to 2} \frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{(\sqrt{3-x}-1)(\sqrt{3-x}+1)}=\lim_{x \to 2} \frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)}{2-x}$$

$$=\lim_{x \to 2} \frac{(\sqrt{6-x}-2)(\sqrt{3-x}+1)(\sqrt{6-x}+2)}{(2-x)(\sqrt{6-x}+2)}=\lim_{x \to 2} \frac{(\sqrt{3-x}+1)(2-x)}{(2-x)(\sqrt{6-x}+2)}=$$

$$\lim_{x \to 2} \frac{\sqrt{3-x}+1}{\sqrt{6-x}+2}=\frac{1}{2}$$

• Ahhh duh.. potato brain moment. In my work, I was calculating the limit directly from the last equality in your before last line and I was getting another indeterminate form... aka wasn't cancelling out my (2-x).... Duh. Thank you! – Sam Belliveau Dec 15 '16 at 19:58
• You are very welcome! – Arnaldo Dec 15 '16 at 20:44

If we put $y=2-x$, then the function becomes

$$\frac{\sqrt{4+y}-2}{\sqrt{1+y}-1}$$

$$=\frac{2(\sqrt{1+\frac{y}{4}}-1)}{\sqrt{1+y}-1}$$

$$2\frac{1+\frac{y}{8}(1+\epsilon_1(x))-1}{1+\frac{y}{2}(1+\epsilon_2(x))-1}$$

thus, the limit when $y$ goes to $0$ is

$$\frac{1}{2}.$$

$$\lim_{x\rightarrow2}\frac{\sqrt{6-x}+2}{\sqrt{3-x}+1} = 2$$

And so: $$\lim_{x\rightarrow2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}=\frac{1}{2}\lim_{x\rightarrow2}\left(\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\cdot\frac{\sqrt{6-x}+2}{\sqrt{3-x}+1}\right)= \frac{1}{2}\lim_{x\rightarrow2}\frac{2-x}{2-x}=\frac{1}{2}$$