Question

Limits questions

Answer 1

I'm working on 21-24. I got 3 as the limit for 21, but I wasn't sure what they meant by "the definition of the derivative". I just multiplied out and simplified, then plugged in 0 for h.

Answer 2

the form of the question is the definition of limit,

Answer 3

\[\lim_{h \rightarrow 0}\frac{ f(x+h) - f(x)}{ h }\]

Answer 4

Definition of Derivative:
\[\lim_{h \rightarrow 0} \frac{ f(x+h) - f(x) }{ h }\]

Answer 5

The numerator is composed of these two pieces:

The first part:\[\large\rm f(\color{orangered}{1+h})=2(\color{orangered}{1+h})^2-(\color{orangered}{1+h})\]And then what is being subtracted off (the 1) is actually this,\[\large\rm f(\color{orangered}{1})=2(\color{orangered}{1})^2-(\color{orangered}{1})\]

Answer 6

HI!!

Answer 7

Do you you sort of see what's going on abble?
This is the limit definition for this function,\[\large\rm f(\color{orangered}{x})=2(\color{orangered}{x})^2-(\color{orangered}{x})\]

Answer 8

they are telling you they are derivatives , that is the hint

Answer 9

evaluated at x=1.

Answer 10

u might also see this as a definition for limit :
\[\left| F(x) - L \right| <\epsilon \]
\[\left| x- a \right| < \delta\]

Answer 11

so for example 22 is the derivative of sine, evaluated at \(\frac{\pi}{2}\)
if you know the derivative of sine, plug in \(\frac{\pi}{2}\)

Answer 12

Yes zep, that makes sense... how would I go about evaluating the limit?

Answer 13

Once you've discovered what the function is,
just use your shortcut for differentiating (power rule)
and then plug in x=1.

Bypass the whole limit thing.

Answer 14

Wait, is the formula that two people posted about the definition of the derivative or the definition of a limit?

Answer 15

The limit definition of a derivative.

Answer 16

Okay, thanks. So when I differentiate I should get 4x - x, plug in 1 and get 3 as the limit :D Thanks zep! What about for the others though? They don't look exactly like the formula...

Answer 17

If \(\large\rm f(x)=2x^2-x\)

Then,\[\large\rm f'(1)=\lim_{h\to0}\frac{2(1+h)^2-(1+h)-\left[2(1)^2-1\right]}{h}\]

Answer 18

Ya, you're taking a shortcut :D Fun stuff!

Answer 19

How bout 23? Think about what's going on with that pi :) hmm

Answer 20

mmmm pie

Answer 21

If it IS the form of a derivative, then it must be something like this,\[\large\rm \lim_{h\to0}\frac{\cos(\pi+h)+1}{h}\quad=\quad \lim_{h\to0}\frac{\cos(\pi+h)-\cos(\pi)}{h}\quad\]right? :o

Answer 22

Ooh interesting

Answer 23

So then is f(x) -cos(pi) ?

Answer 24

It looks like f(x)=cos(x) right?
Evaluated at x=pi

Answer 25

Ohh okay

Answer 26

\(\large\rm f(x)=cos(x)\)

With,\[\large\rm f'(\pi)=\lim_{h\to0}\frac{\cos(\pi+h)-\cos(\pi)}{h}\]

Answer 27

Okay that makes perfect sense. Thanks zep. I see 22 now as well. f(x) will be sin(x) evaluated at pi/2.

Answer 28

What about that 24 though O.O

Answer 29

Oh 24.. they're using the OTHER derivative definition.
Do you remember this one?\[\large\rm f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\]It doesn't come up as often.

Answer 30

\[\large\rm f'(8)=\lim_{x\to 8}\frac{f(x)-f(8)}{x-8}\]

Answer 31

Oh yeah, that does look familiar. Would I find the derivative of f(x)? Which would be 2x^(-1/3) I believe

Answer 32

Evaluated at x=8, which would give me 1 as the limit

Answer 33

Hmm I dunno.. I think that 3 is messing things up...

Answer 34

https://www.wolframalpha.com/input/?i=limit+as+x+approaches+8+of+(3x%5E(2%2F3)-4)%2F(x-8)

Answer 35

Ya someone made a typo here.
Perhaps they meant for the problem to be this,\[\large\rm \lim_{x\to8}\frac{3(x^{2/3}-4)}{x-8}\]Because it doesn't make any sense right now... hmm

Answer 36

Because if f(x) = 3x^(2/3)
Then we would expect f(8) = 3(8)^(2/3)

But that doesn't give us 4.

Answer 37

Do you have an answer key to go along with these? :d

Answer 38

The answer key says 1... hmm

Answer 39

\[\large\rm \lim_{x\to8}\frac{3(x^{2/3}-4)}{x-8}=1\]Ya just a typo :d
Or more likely the teacher probably intended for it to be this,\[\large\rm \lim_{x\to8}\frac{3x^{2/3}-12}{x-8}\]Which is the same thing.

Answer 40

Thanks so much zep :)

Answer 41

Typos are the worst in math tho :/