Find the limit of the this equation

[shwanky]

Homework Statement

Evaluate the limit if it exists.

Homework Equations

lim_{x\to2} \frac{\sqrt{x-6}-2}{\sqrt{x-3}-1}

The Attempt at a Solution

Completely lost, I've tried, finding the conjugate of both the numerator or denominator and I'm unable to find a solution.

 

[HallsofIvy]

Are you sure that is the problem? AT x= 2, the fraction is
\frac{2i- 2}{i- 1}= 2
Nothing hard about that!

Perhaps you meant
\frac{\sqrt{6- x}- 2}{\sqrt{3- x}-1}[/itex]<br /> which is &quot;0/0&quot; when x= 2.<br /> Use the &quot;conjugates&quot; as you mention- multiply both numerator and denominator by (\sqrt{x-6}+2)(\sqrt{x-3}+1). You will get x²-4 in the numerator and x-2 in the denominator and can cancel.

 

[shwanky]

ok, here's another one I don't understand.

Homework Equations

lim_{t \to 0} \frac{x+1}{xsin(\Pi x)}

The Attempt at a Solution

lim_{t \to 0} \frac{x+1}{xsin(\Pi x)}(\frac{x-1}{x-1})
lim_{t \to 0} \frac{(x^2-1)}{(x^2-1)sin(\Pi x)}
lim_{t \to 0} \frac{(1)}{sin(\Pi x)}

At this point I get stuck... the sin(\Pi x) is still 0. I tried using some of the trig identities, that I actually remembered, and got this.

lim_{t \to 0} \frac{(sin^2(\Pi x) + cos^2(\Pi x)}{sin(\Pi x)}

Now I thought if I could get I could get the numerator in terms of sine, I could simple divide into it, but I'm not sure if this would work :-/... any suggestions?

 

[shwanky]

is it possible for me to do

lim_{t \to 0} \frac{(1)}{sin(\Pi x)}
lim_{t \to 0} \frac{(1)}{csc(\Pi x)}
lim_{t \to 0} sin(\Pi x)

?

 

[shwanky]

mmm never mind. there was a problem in my solution. xsin(\Pi x)(x-1) does not equal (x2-1)sin(\Pi x)... I will try again...

 

[shwanky]

ok, I think I got it.

lim_{t \to 1} \frac{x+1}{xsin(\Pi x)}

lim_{t \to 1} \frac{x+1}{xsin(\Pi x)}(\frac{csc(\Pi x)}{csc(\Pi x)})

lim_{t \to 1} \frac{(x+1)(csc(\Pi x)}{x}

lim_{t \to 1} \frac{(x+1)(csc(\Pi x)}{x}

lim_{t \to 1} csc(\Pi x) + \frac{(csc(\Pi x)}{x}

lim_{t \to 1} csc(\Pi * 1) + \frac{(csc(\Pi *1)}{1}

lim_{t \to 1} csc(\Pi) + (csc(\Pi)

lim_{t \to 1} 2csc(\Pi)

lim_{t \to 1+} 2csc(\Pi) = +\infty

lim_{t \to 1-} 2csc(\Pi) = -\infty

 

[shwanky]

Sorry, the limit is t->1+ not t->0...

 

[shwanky]

ok, now if I take the left hand limit and right hand limit, the right hand limit approaches infinity while the left hand appraches negative infinity?

 

[HallsofIvy]

Does it bother you at all that you are taking the limit as t-> 1 but there is no t in the formula?! (Mathematics requires precision- be careful what you write!)

Perhaps you meant
lim_{t \to 1^+} \frac{t+1}{tsin(\Pi t)}

I don't see difficulty with that. The denominator is continuous and at t= 1 is (1)sin(\pi)= 0 but the numerator does NOT go to 0. What does that tell you?

(Don't capitalize "Pi"- you want \pi: \pi.)

 

[shwanky]

... I am such a dork...

\lim_{t \to 1^+} \frac{t+1}{tsin(\pi t)} = -\infty as t \to 1^+