Indeterminate Limits: Algebraically

[TylerH]

I had a test earlier today, and there were a few problems I had no clue as to how to answer. I'm a perfectionist, so I want to learn what I did wrong, and what to do differently if I ever come along a similar problem again.

One of the problems I couldn't do was to find $$lim_{x \to 16} \frac{4 - \sqrt{x}}{x - 16}$$. I vaguely remember that I am supposed to multiply by the conjugate $$\frac{x+16}{x+16}$$, but then what?

All the others were similar to the above, in that it was trivial to get a diff of squares on the bottom, I just couldn't remember what to do after I got the diff of squares on the bottom.

 

[mjpam]

Are you sure that you want to multiply by $\frac{x+16}{x+16}$?

Does doing so eliminate the division by 0 as $x\to16$?

 

[mjpam]

Put another way: Would you know how to solve the problem if it were
$$lim_{x\to2}\frac{4-x^{2}}{x-2}$$

 

[Hurkyl]

I'm a perfectionist, so I want to learn what I did wrong, and what to do differently if I ever come along a similar problem again.

Ah, but you failed to ask the right question: "How does multiplying by the conjugate help simplify an expression?"

 

[HallsofIvy]

I would rather multiply both numerator and denominator by $4+ \sqrt{x}$.

 

[mjpam]

I would rather multiply both numerator and denominator by $4+ \sqrt{x}$.

Wouldn't factoring the denominator work better?

 

[TylerH]

Sorry it took so long. We had Saturday school here, to make up for snow days, and I've been doing hw.

Taking HallsofIvy's suggestion:
$$lim_{x \to 16} \frac{4 - \sqrt{x}}{x - 16}$$
$$lim_{x \to 16} \frac{(4 - \sqrt{x})(4 + \sqrt{x})}{(x - 16)(4 + \sqrt{x})}$$
$$lim_{x \to 16} \frac{(x - 16)}{(x - 16)(4 + \sqrt{x})}$$
$$lim_{x \to 16} \frac{1}{(4 + \sqrt{x})}=\frac{1}{8}$$

Was it that I had the rule wrong, or that this problem isn't of the same class of problem where you would need to multiply by the conjugate of the denominator?

mpjam, how would you do it the factoring way? (x-16) factors to (sqrt(x)+4)(sqrt(x)-4), which doesn't cancel with anything AFAIK.

 

[spamiam]

$$\sqrt{x} - 4 = -(4 - \sqrt{x})$$

So there is some cancellation. But more importantly, check your multiplication again. I think you dropped a minus sign.

 

[TylerH]

Oh, duh! I'm so used to seeing diff of squares with x as the positive, that as soon as I saw them, I assumed it would be the same. But you're right, it would be 16-x, rather than x-16.