Can conjugates be used to evaluate this limit?

[eric_f]

Hi All,

I ran into a stumper of a limit problem that requires more brainpower than I have at the moment.

Evaluate the following limit:

\(\displaystyle \lim_{x\rightarrow 4} \frac{\sqrt{5-x}-1}{2-\sqrt{x}}\)

Thus far I've tried multiplying by the conjugate of the numerator and the denominator with no foreseeable cancellations or beneficial groupings. Can someone give me a push in the right direction?

I end up with either:

\(\displaystyle \lim_{x\rightarrow 4} \frac{4-x}{2\sqrt{5-x}+2-\sqrt{x}\sqrt{5-x}-\sqrt{x}}\)

or

\(\displaystyle \lim_{x\rightarrow 4} \frac{2\sqrt{5-x}+\sqrt{x}\sqrt{5-x}-2-\sqrt{x}}{4-x}\)


and grind to a screeching halt...I'm hoping it's something simple




Thanks!

 

[DrPhil]

Hi All,

I ran into a stumper of a limit problem that requires more brainpower than I have at the moment.

Evaluate the following limit:

\(\displaystyle \lim_{x\rightarrow 4} \frac{\sqrt{5-x}-1}{2-\sqrt{x}}\)

Thus far I've tried multiplying by the conjugate of the numerator and the denominator with no foreseeable cancellations or beneficial groupings. Can someone give me a push in the right direction?

I end up with either:

\(\displaystyle \lim_{x\rightarrow 4} \frac{4-x}{2\sqrt{5-x}+2-\sqrt{x}\sqrt{5-x}-\sqrt{x}}\)

or

\(\displaystyle \lim_{x\rightarrow 4} \frac{2\sqrt{5-x}+\sqrt{x}\sqrt{5-x}-2-\sqrt{x}}{4-x}\)


and grind to a screeching halt...I'm hoping it's something simple
Thanks!

Maybe the complexity is telling you something - like "That ain't it."

I am hoping that you have learned l'Hospital's rule - this turns out to be a good time to use that rule!

 

[eric_f]

Maybe the complexity is telling you something - like "That ain't it."

I am hoping that you have learned l'Hospital's rule - this turns out to be a good time to use that rule!

This book (Stewart, 2008) doesn't introduce that until Chapter 7 unfortunately, and I'm just at the end of 2! As a matter of fact, I don't even see it on the syllabus so it may very well be Calculus 2 material... This isn't an assigned question, but I figured I would give it a go anyway. Serves me right for being curious If l'Hospital's rule isn't applied, how convoluted of a process would one go through to evaluate this problem?

 

[DrPhil]

This book (Stewart, 2008) doesn't introduce that until Chapter 7 unfortunately, and I'm just at the end of 2! As a matter of fact, I don't even see it on the syllabus so it may very well be Calculus 2 material... This isn't an assigned question, but I figured I would give it a go anyway. Serves me right for being curious If l'Hospital's rule isn't applied, how convoluted of a process would one go through to evaluate this problem?

Let \(\displaystyle f(x) = \sqrt{5 - x} - 1\)

......\(\displaystyle g(x) = 2 - \sqrt{x}\)

\(\displaystyle \displaystyle \lim_{x\to4} \frac{f(x)}{g(x)} = \lim_{h\to0} \frac{f(4+h)}{g(4+h)} = \lim_{h\to0} \frac{f(4)+h\ f'(4)}{g(4) + h\ g'(4)}\)

But since f(4) = 0 and g(4) = 0, and h/h cancels to 1, this becomes

\(\displaystyle \displaystyle \lim_{x\to4} \frac{f(x)}{g(x)} = \dfrac{f'(4)}{g'(4)}=\dfrac{\lim_{h\to0}[f(4+h) - f(4)]/h}
{\lim_{h\to0} [g(4+h) - g(4)]/h}\)

 

[lookagain]

Multiply the numerator and the denominator by the product of the conjugates of the numerator and denominator. Then reduce that fraction by cancelling out the (4 - x) terms from the new numerator to the new denominator.

\(\displaystyle \lim_{x\rightarrow 4} \frac{\sqrt{5 \ - \ x} \ - \ 1}{2 \ - \ \sqrt{x}} \ = \)



\(\displaystyle \lim_{x\rightarrow 4} \frac{2 \ + \ \sqrt{x \ }}{\sqrt{5 \ - \ x \ } \ \ + \ 1}\)

.