Finding Limits with a Constant: Tips for Calculus Students

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Homework Statement



Find in terms of the constant a

Lim h --- > 0

8 (a+h)² - 8a²
-----------------
h


The Attempt at a Solution



I have no idea how to even approach it so some pointers would be nice.
It hasn't been long since I started calculus. I am not sure weather we covered this
or not , classes are so heavy with content and passes quickly :s .

anyways what I am sure we have seen until now is

-IVT
- squeeze
- Limit laws

we have seen some things concerning continuity theorems altho am currently lost in them.
 
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Well,if you substitute in the value of h=0 directly, then you will get 0/h = 0.
So, why don't you try to expand the numerator? Then, proceed to the operations and eventually, the h at the denominator will be canceled out.
 
well I end up with 16a = something . am giong to look at it more
 
Last edited:
huh.. well the answer Is 16a altho I don't really understand it.
 
Ooo I get it now well thanks for your time.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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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