Limits and finding a constant value 'k'

Cpie05
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Stick with me here, I don't know how to use something to add an equation in here!

Homework Statement



Find a value of the constant k such that the limit exists:

lim x->4 (x^2 - k^2) / (x-4)


The Attempt at a Solution



I KNOW the solution is the limit will exist iff k = -4 and k = 4.

My problem is - does the numerator have to cancel out the denominator in order for a limit to exist? Or is that just the case here?

Cause the solution is:

(x-k)(x+k)/(x-4)

So the only way for the denominator (x-4) to cancel out would be if k = -4 or 4... so I'm just wondering if that's a general rule?

Cheers
C.
 
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The rule is that if the denominator goes to zero, the only way you can have a limit is if the numerator also goes to zero. For reasons that should be obvious if you think about how division works.
 
Go back and review the limit theorems. Pay particular attention to the details on the one for division; a surprising number of people completely ignore them, and then have trouble dealing with limits that have a division in them. :frown:

By applying the fact the limit exists, you should be able to determine something about the thing of which you're taking the limit.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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