Question

Use graphs and tables to find the limit and identify any vertical asymptotes of the function.

limit of 1 divided by the quantity x minus 1 squared as x approaches 1

Answer 1

is this right?
lim x->1 = f(x+h)−f(x)/h
lim x->1 = −4−2x−2h−(−4−2x)/h
lim x->1 = -2h/2
= -2
@ganeshie8 @Hero @TuringTest @iambatman

Answer 2

that was the wrong question for my answer! sorry here's the real question!
The position of an object at time t is given by s(t) = -4 - 2t. Find the instantaneous velocity at t = 6 by finding the derivative.

Answer 3

well I also need help on the question I posted accidently... so if someone could help me on that I would really appriceiate it!:) and I know what the vertical asymptote is but how do I find the limits?

Answer 4

I feel like I confused you guys so here is a clarification!:)
1. The position of an object at time t is given by s(t) = -4 - 2t. Find the instantaneous velocity at t = 6 by finding the derivative.
and my answer to this...
lim x->1 = f(x+h)−f(x)/h
lim x->1 = −4−2x−2h−(−4−2x)/h
lim x->1 = -2h/2
= -2
and is this correct?

2. Use graphs and tables to find the limit and identify any vertical asymptotes of the function.

limit of 1 divided by the quantity x minus 1 squared as x approaches 1
** I know the vertical asymptote to this is 1, but how do I find the limits to this?

Answer 5

@ParthKohli @bohotness @abb0t @mathmate @LazyBoy @Hero @dan815 @perl someone please help?:)

Answer 6

@pizza12

Answer 7

@some.random.cool.kid can you pretty please help me??:)

Answer 8

looks fine :}

Answer 9

@some.random.cool.kid do you know how to find the limits in my second question?

Answer 10

@bohotness

Answer 11

@uri

Answer 12

is the limit -infinity?

Answer 13

yes.

Answer 14

1. velocity : your answer is correct
s(t)=-4-2t
s'(t)=-2
therefore instantaneous velocity at any time is -2 units.

2. \(Lim_{x\rightarrow 1} \dfrac{1}{(x-1)^2}\)
As you have correctly pointed out, the vertical asymptote is at x=1, because the denominator becomes zero at that point.
If you want to go a little further, you can sketch the graph. Doing so will help you visualize other problems easily.
To find the limit, it is often easoer to start by substitute the limiting value of x. If you do not get 0/0, or \(inf /inf\), then the evaluated expression is usually the limit.
When the value of the expression is infinity, we could say that the limit does not exist, or as you said, it's infinity.

Answer 15

The curve is not supposed to touch the vertical line x=1. (bad drawing!)

Answer 16

awesome! thanks so much for your help:))

Answer 17

You're welcome! :)