Question

lim2x-x^2/2-x

Answer 1

lim as x approaches what?

Answer 2

2

Answer 3

\[\text{this }\lim_{x\to 2}\frac{2x-x^2}{2-x}\text{ ?}\]

Answer 4

yes

Answer 5

you really need to use parentheses ;)

Answer 6

then I think the answer is 2. Here's my take on it...
the numerator can be factored as x (2-x) hence x(2-x)/(2-x)
we see that the 2-x will cancel to leave x in the numerator. Substitute 2 in x and the result is 2

Answer 7

Thanks.

Answer 8

What about this one? lim\[2^{1/x}\] as x approaches infinity?

Answer 9

If I am not mistaken then the limit should be 1. My reasoning is that as x approaches infinity then1/x approaches 0 and anything number raised to the power of 0 is 1.

Answer 10

why does 1/x approach 0 as x approaches infinity?

Answer 11

think of it as dividing 1 by a very large number. 1/1000000000000<1/10000<1/100<1/10........so we see that the larger the number 1 is divided by, the smaller its value. As x (the number we divide by) gets infinitely large then the value of 1/x approaches 0.

Answer 12

Thank you so much :)

Answer 13

Cheers!!

Answer 14

\[\lim_{ \rightarrow \infty} (\sqrt{{x+1}}\]

Answer 15

oops, -\[\sqrt{x}\]

Answer 16

the first post minus the second

Answer 17

Can you rewrite it again. Am not sure what the question is...

Answer 18

\[\lim_{ \rightarrow \infty}(\sqrt{x+1} -\sqrt{x})\]

Answer 19

On a first look I am tempted to say that this approaches infinity, but upon closer scrutiny it seems like there is something more subtle going on here because \[\infty-\infty seems rather indeterminate\]. Whenever I approach a problem like this I try to rewrite it by multiplying by 1.ie use a clever way to rearrange the problem and since multiplying by 1 doesn't change its value, let's give it a shot. I am thinking of multiplying the above expression by it's conjugate (\[\sqrt{x+1}+\sqrt{x}\]

so \[\lim_{x \rightarrow \infty}(\sqrt{x+1}-\sqrt{x})*((\sqrt{x+1}+\sqrt{x}))/(\sqrt{x+1})+\sqrt{x})\]

This gives: \[[(x+1)-x]/[\sqrt{x+1}+\sqrt{x}]=1/[\sqrt{x+1}+\sqrt{x}] \]
So as x approaches infinity the denominator approaches \[\infty\] and we are back to that 1/\[\infty\] expression which indicates that the limit is 0. let us know if this is correct.

Answer 20

Okay, thanks again.

Answer 21

No worries! All the best with your studies