Limit + Derivatives

[funnytim]

Hey everyone,

I have some questions on finding the limits:


(31)
lim (x-4)
x->4 (square root x) -2

i started off by squaring the numerator + denominator, resulting in:
x^2-8x+16
x-4(square root x) + 4

But now what can i do? If I plug in x =4 into the eqn, it is undefined which isn't the answer (ans=4)


(27)
In the below question I'm supposed to use the 'squeeze law of limits to find the limits'. But I don't understand exactly how to use it
lim x^2 cos (1 / (cube root x)
x->0
Ans = 0


How can I find out whether the limit's discontinuity is removable or not? eg,
x-17
[abs(x-17)]


Lastly, I sort of blanked out, how do you find the derivative of a root, or fraction? Egs:
Find f ' (x) of (square root 2x+1) , x / (1-2x)


Thanks everyone!

 

[galactus]

Hey everyone,

I have some questions on finding the limits:

(31)
lim (x-4)
x->4 (square root x) -2

i started off by squaring the numerator + denominator, resulting in:
x^2-8x+16
x-4(square root x) + 4

But now what can i do? If I plug in x =4 into the eqn, it is undefined which isn't the answer (ans=4)

Just use a little algebra(long division) and it whittles down to \(\displaystyle \sqrt{x}+2\). Now, plug in x=4 and we get 4 as the limit.

 

[BigGlenntheHeavy]

How can I find out whether the limit's discontinuity is removable or not? eg,
x-17
[abs(x-17)]


\(\displaystyle Hint: \ \lim_ {x\to17^{-}}\frac{x-17}{|x-17|}} \ = \ -1 \ and \ \lim_{x\to17^{+}}\frac{x-17}{|x-17|} \ = \ 1\)

\(\displaystyle Does \ that \ tell \ you \ anything?\)

\(\displaystyle One \ can \ remove \ a \ point, \ but \ a \ gaping \ gash, \ I \ don't \ think \ so.\)

 

[funnytim]

Hey everyone,

I have some questions on finding the limits:

(31)
lim (x-4)
x->4 (square root x) -2

i started off by squaring the numerator + denominator, resulting in:
x^2-8x+16
x-4(square root x) + 4

But now what can i do? If I plug in x =4 into the eqn, it is undefined which isn't the answer (ans=4)

Just use a little algebra(long division) and it whittles down to \(\displaystyle \sqrt{x}+2\). Now, plug in x=4 and we get 4 as the limit.

As in, do long division on:
x^2-8x+16
x-4(square root x) + 4 ?

How can I find out whether the limit's discontinuity is removable or not? eg,
x-17
[abs(x-17)]


\(\displaystyle Hint: \ \lim_ {x\to17^{-}}\frac{x-17}{|x-17|}} \ = \ -1 \ and \ \lim_{x\to17^{+}}\frac{x-17}{|x-17|} \ = \ 1\)

\(\displaystyle Does \ that \ tell \ you \ anything?\)

\(\displaystyle One \ can \ remove \ a \ point, \ but \ a \ gaping \ gash, \ I \ don't \ think \ so.\)

So in the two limit eq'ns, I 'plug in' x = 17 and x=-17 to find that the two limits don't equal each other, so its non-removable?


Thanks again!