Question

l'Hospital's rule: sin(x)/3x

Answer 1

i need to find limits!

Answer 2

as x approaches 0?

Answer 3

yeah

Answer 4

You don't need L'Hop for this one.
Did they instruct you to use it though? :)

Answer 5

well they said if we didn't have then we don't but we are encouraged to? Can you show me both ways?

Answer 6

no need

Answer 7

so do you get the derivatives of both? if I was to use LHop:?

Answer 8

The first way is to remember a really important identity, it shows up a lot to try to commit it to memory.\[\large \lim_{x \to 0}\frac{\sin x}{x}=1\]

Answer 9

\[\lim_{x\to 0}\frac{1}{3}\frac{\sin(x)}{x}\]

Answer 10

\[=\frac{1}{3}\lim_{x\to 0}\frac{\sin(x)}{x}\]

Answer 11

ok~

Answer 12

what @zepdrix said finishes it

Answer 13

so if I was to use L'hop rule????

Answer 14

Ok let's see :)

Answer 15

or what happens to the 1/3? when the method you showed me is used?

Answer 16

it just sits there

Answer 17

lol so the final answer is?

Answer 18

\[\lim_{x\to a}cf(x)=c\lim_{x\to a}f(x)\]

Answer 19

\[\large \color{orangered}{\lim_{x \to 0}\frac{\sin x}{x}=1}\]


\[\large \frac{1}{3}\color{orangered}{\lim_{x\to 0}\frac{\sin(x)}{x}} \qquad=\qquad \frac{1}{3}\color{orangered}{1}\]

Answer 20

ahh...

Answer 21

oooh orange is the new black!

Answer 22

XD

Answer 23

\[\large \lim_{x \to 0}\frac{\sin x}{3x}\qquad\rightarrow\qquad \frac{0}{0}\]

This limit is approaching 0/0, which IS one of the indeterminate forms we're allowed to apply L'Hopital's Rule to!

So taking the derivative of the top and bottom (separately) gives us,\[\large L'H \qquad=\qquad \lim_{x \to 0}\frac{\cos x}{3}\]

Answer 24

right.

Answer 25

So what happens as x approaches 0 now? :O hmm

Answer 26

and cos(x) equals one so 1/3?

Answer 27

yay good job \:D/

Answer 28

cos(0)=1, yes

Answer 29

ok can we work on one more?

Answer 30

so I can make sure I got this?

Answer 31

sure :3

Answer 32

cos(x)/ sinx+1

Answer 33

do we need to use LHop?

Answer 34

x approaching?

Answer 35

oh zero lol

Answer 36

Hmmm.. Can't we just evaluate this one directly?
Plugging in x=0 gives us,\[\large \frac{\cos0}{\sin0+1} \qquad=\qquad \frac{1}{0+1}\]

Answer 37

i don't know I guess...that'll be ok? hopefully?

Answer 38

Well I worked it out with L'Hopital and it's giving us the same answer.
So yah.. it should be ok.

Not sure why you were given an easy problem like that.
Maybe just to confuse you lol.

Answer 39

hmmm ok! ONE MORE! just one more?

Answer 40

fine fine fine XD

Answer 41

xe^-x

Answer 42

as x approaches infinity

Answer 43

hmph i hate e's

Answer 44

Ok so you'll want to make sure you get comfortable with exponents.
This can be a little tricky to evaluate directly if you don't remember how negative exponents work.

Let's look at the exponential a sec,

\(\large \lim_{x\to\infty}e^{-x}\qquad\to\qquad0\)

Understand why this is approaching 0?

Answer 45

no. lol

Answer 46

i thought it was approaching infinity?

Answer 47

oh shoot I gotta run! but can you please explain it to mee and I'll def check it when I get back??

Answer 48

Let's plug in a really really big value (close to infinity) for x and see what happens.

\[\huge e^{-99999} \qquad=\qquad \frac{1}{e^{99999}} \qquad\approx\qquad 0\]Remember how the negative on the exponent tells us to flip it to the denominator?
So this gives us 1/(a bajillion), which is really really close to zero.

Answer 49

fine fine fine, ill leave some notes -_-

Answer 50

I REALLY like how your explaining it!!!

Answer 51

hahaa thanks man

Answer 52

ohhh and i got ur last note. I understand why its approaching zero

Answer 53

\[\large \lim_{x\to\infty}xe^{-x} \qquad\to\qquad \infty\cdot0\]
This is an indeterminate form, but it's NOT in the proper form for us to apply L'Hopital's Rule.

So like I did in that example with the negative exponent, let's do it here also.\[\large \lim_{x\to\infty}xe^{-x} \qquad=\qquad \lim_{x\to\infty}\frac{x}{e^x} \qquad\to\qquad \frac{\infty}{\infty}\]
Ok good! Now it's giving us the form we need for L'Hopital.

Answer 54

Applying L'Hopital's Rule, separately taking the derivative of the top and bottom, gives us,\[\large L'Hop \qquad=\qquad \lim_{x\to\infty}\frac{1}{e^x}\]

Answer 55

If we want to, we could rewrite this as,\[\large \lim_{x\to\infty}\frac{1}{e^x}\qquad=\qquad\lim_{x\to\infty}e^{-x}\]

And I think we already talked about this particular limit before :)
So just scroll up if you forgot. ( The thing with the 1/[a bajillion] ).

Answer 56

again?

Answer 57

haha i just saw the reply from the other person!