Question

Calculus help!

Answer 1

@Zarkon

Answer 2

Do you understand what is going on with something like this? :)\[\large\rm \lim_{x\to0^+}\frac{1}{x}\]

Answer 3

no

Answer 4

Well i took the derivvative of the equation and got 0. So would A be the answer?

Answer 5

Sorry I ran off for a sec ^^

Answer 6

Let's examine this example I posted really quickly,
should lead you to the correct answer.

Answer 7

If x is approaching 0 from the right,
let's plug in a number which is really really really close to 0 and get an idea of what is going on.

Answer 8

So here is a number very close to zero, 1/9999999

Answer 9

\[\large\rm \lim_{x\to0^+}\frac{1}{\color{orangered}{x}}\approx\frac{1}{\left(\color{orangered}{\frac{1}{99999999}}\right)}\]We're dividing by a fraction, so we can rewrite it like this,\[\large\rm \lim_{x\to0^+}\frac{1}{\color{orangered}{x}}\approx1\cdot \frac{99999999}{1}\]

Answer 10

So you can see that as x gets smaller and smaller,
the value of this fraction gets larger and larger, ya?

Answer 11

As \(\rm x\to0^+\), we have \(\rm \frac{1}{x}\to\infty\)

Answer 12

\[\large\rm \lim_{x\to0^+}\left(\frac{7}{x}-\frac{3}{x^2}\right)\quad=\infty-\huge \infty\]This might seem a little sloppy, your teacher probably wouldn't accept this type of notation,
but notice the x is being squared on the second term.
So that one is actually blowing up much faster.

Answer 13

okay, that makes sense. so it would just be infinity?

Answer 14

Well if you have a number minus a larger number,
then it's negative, ya?

I'm trying to remember the more appropriate way to do this.. hmm
Oh oh here we go.
Let's factor a 1/x^2 out of each term.

Answer 15

\[\large\rm \lim_{x\to0^+}\frac{1}{x^2}\left(7x-3\right)\]Any confusion on that step?

Answer 16

Negative, correct. and no

Answer 17

\[\large\rm \lim_{x\to0^+}\frac{1}{x^2}\left(7x-3\right)\quad=\infty(0-3)\]So I guess we see that it's approaching something like this.

Answer 18

So negative infinity, ya :)

Answer 19

Right, thank you!!!