# Find the limit of the this equation

## 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
Homework Helper
Are you sure that is the problem? AT x= 2, the fraction is
$$\frac{2i- 2}{i- 1}= 2$$

Perhaps you meant

## 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?

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is it possible for me to do

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

?

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...

ok, I think I got it.

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

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

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

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

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

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

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

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

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

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

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

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
Homework Helper
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: $\pi$.)

... I am such a dork...

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

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