Question

L'Hopitals rule: evaluate the limit:

Answer 1

lhops rule works for limits that give us 0/0 or inf/inf, if memory serves

Answer 2

\[\lim_{x \rightarrow 0} \frac{ \sin^2 x }{ \tan(x^2) }\]

Answer 3

right.

Answer 4

so the first thing i think I'd do is take the derivative of the top and the bottom, because if i just try plugging in 0 i'll get 0 in the denominator

Answer 5

you can apply the L'H'S now

Answer 6

Hints:
\(\large\color{black}{\sin^2x=(\sin~x)^2}\) and then use the chain rule.
for \(\large\color{black}{\tan(x^2)}\) you will need a chain rule also.

Answer 7

hey guys, let him try

Answer 8

right, so i got. \[\frac{2 \sin x \cos x}{2xsec^2 x^2}\]

Answer 9

\[\sin^2(x) +\cos^2(x) = 1\]For \(\sin^2(x)\) , can you not replace it with \(1-\cos^2(x)\)?

Answer 10

L'Hospitals rule is overkill for this problem

Answer 11

the derivative is correct, study guy

Answer 12

okay.

Answer 13

I would like to hear another way to do it asides from L'h'S if there is any. I am not good at mathematics, and I might miss a better approach sometimes.

Answer 14

hospitalization may be overkill, but I think the purpose of the question is to get someone used to the concept

Answer 15

hospitalization LOL

Answer 16

okay so you can replace the sin^2 x with 1 - cos^2 x because of trig identites

Answer 17

I mean, it sounds like you HAVE TO use it, so we will go that way.

Answer 18

I just wish they gave problems to students that necessitated the need for L'Hospitals rule

Answer 19

Yeah, before taking derivatives I would always reference trig identities first so you can simplify the function before using any simplification rules.

Answer 20

it's like asking someone to evaluate using limit definition... it is overkill, but it serves a purpose

Answer 21

Well, anyway, it is a requirement so lets proceed.

We got: \(\Large\color{black}{\displaystyle\lim_{x \rightarrow ~a} ~\frac{ 2\sin(x)\cos(x)}{ 2x\sec^2(2x)}}\)

Answer 22

Okay, when you plug in I mean not a,

\(\Large\color{black}{\displaystyle\lim_{x \rightarrow ~0} ~\frac{ 2\sin(x)\cos(x)}{ 2x\sec^2(2x)}}\)

Answer 23

if you plug in zero into the limit, what do you get, study guy?

Answer 24

well the denominator is 0 again. so we use lhopital's again?

Answer 25

No no

Answer 26

if the denominator and numerator is zero.

Answer 27

and this is the case, but I just want to tell the concept, that if you get (for example) 1/0, you can NOT apply L'H'S.

Answer 28

but in this case it is still 0/0, right because sin(0) is zero, and the 2x is zero, when x=0.

Answer 29

\(\Large\color{black}{\displaystyle\lim_{x \rightarrow ~0} ~\frac{ 2\sin(x)\cos(x)}{ 2x\sec^2(2x)}}\)

Answer 30

Apply L'H's rule again.

Answer 31

okay

Answer 32

take your time:)

Answer 33

should i first use the trig identity to make the 2sin(x)cos(x) into sin(2x) so it's easier?

Answer 34

Yes!!

Answer 35

yes, if you want.

Answer 36

\(\Large\color{black}{2\sin(x)\cos(x)=\sin(x)\cos(x)+\sin(x)\cos(x)=?}\)

Answer 37

@SolomonZelman keep it simple!!

Answer 38

Am I not?

Answer 39

Lol.. taking the derivative of \(2x\sec^2(2x)\) is also a headache enough.

Answer 40

okay so i think i got it is it:
\[\frac{ 2\cos(2x )}{ 4\sec(x^2)\sec(x^2)\tan(x^2) }\]

Answer 41

not really. taking a derivativ is never a headache. A couple of chains, even I can do that.

Answer 42

\[\Large\color{black}{\displaystyle\lim_{x \rightarrow ~0} ~\frac{ 2\sin(x)\cos(x)}{ 2x\sec^2(2x)}}\]

\[=\Large\color{black}{\displaystyle\lim_{x \rightarrow ~0} ~\frac{ \sin(x)}{ x}\lim_{x \rightarrow ~0}\frac{\cos(x)}{\sec^2(2x)}}\]

Answer 43

^ That's what I was getting at.

Answer 44

oh yeah because of the product rule of limits.

Answer 45

And before learning LH rule, you should already know what the identity \(\dfrac{\sin(x)}{x}\) gives you.

Answer 46

yes, but I thought LHS was required.

Answer 47

you already used L'Hospitals rule once

Answer 48

Sometimes when you are able to spot an identity, you can avoid using LH rule.

Answer 49

true:) But you know how some teacher insist on re-using it again and again whether it's easy or not... although your way really seems convincing and (almost) the only...

Answer 50

The purpose of LH rule is to simplify a function and bring it out of an indeterminant form. Once it is out of indeterminant form,you can do whatever you want. It's a free country.

Answer 51

:P

Answer 52

sure:)

Answer 53

we're left with cos x over sec squared 2x since sinx over x is 1 right?

Answer 54

yes.

Answer 55

\(\Large\color{black}{\displaystyle\lim_{x \rightarrow ~0} ~\frac{ 2\sin(x)\cos(x)}{ 2x\sec^2(2x)}}\)


\(\Large\color{black}{\displaystyle\lim_{x \rightarrow ~0} ~\frac{ \sin(x)\cos(x)}{ x\sec^2(2x)}}\)


\(\Large\color{black}{\displaystyle\lim_{x \rightarrow ~0} ~\frac{ \sin(x)}{ x}\times }\)\(\Large\color{black}{\displaystyle\lim_{x \rightarrow ~0} ~\frac{ \cos(x)}{ \sec^2(2x)}}\)

Answer 56

this is what we have done.

Answer 57

And then, as you said.

\(\Large\color{black}{1\times }\)\(\Large\color{black}{\displaystyle\lim_{x \rightarrow ~0} ~\frac{ \cos(x)}{ \sec^2(2x)}}\)

Answer 58

so now i j just plug in 0 right?

Answer 59

yes, everything is correct. I am just again reposting the work, just to make sure you have all the information and stpes you need to have.

Answer 60

yes, plug in 0

Answer 61

so i got 1 as my limit.

Answer 62

Yup!

Answer 63

the answer is 1

Answer 64

Good job.

Answer 65

I still think it is better to do the following

\[\frac{ \sin^2 x }{ \tan(x^2) }\]

\[=\frac{\sin(x)}{x}\frac{\sin(x)}{x}\frac{x^2}{\tan(x^2)}\]
\[=\frac{\sin(x)}{x}\frac{\sin(x)}{x}\cos(x^2)\frac{x^2}{\sin(x^2)}\to1\times 1\times1\times1=1\]

Answer 66

it's definitely IS better to do that.

Answer 67

anyways, just a little attachment.

Answer 68

save L'Hospitals rule for hard problems

Answer 69

Ugh, accidentally deleted my post.

Answer 70

\[\tan(x^2)=\frac{\sin(x^2)}{\cos(x^2)}\]

Answer 71

\[\lim_{x\to 0}\frac{\sin(x^2)}{x^2}=\lim_{u\to 0}\frac{\sin(u)}{u}=1\]

\[u=x^2\]

Answer 72

oh I got it.

Answer 73

I think it comes not from substitution that \(\large\color{black}{\displaystyle\lim_{x \rightarrow ~0}\frac{\sin^2 x}{x^2}=0}\).


\(\large\color{black}{\displaystyle\lim_{x \rightarrow ~0}\left(\begin{matrix} \frac{\sin x}{x} \\ \end{matrix}\right)^2=0}\)

\(\large\color{black}{\left(\begin{matrix}\displaystyle\lim_{x \rightarrow ~0} \frac{\sin x}{x} \\ \end{matrix}\right)^2=0}\)

Answer 74

just by applying basic limit rules. I meean sub is shorter, but rules are more basic.

Answer 75

either way...

Answer 76

those limits are 1 not 0

Answer 77

yes 1... by bad, I left a zero from the latex that I copied, sorry.

Answer 78

(but you get what I was trying to say)

Answer 79

yeah,this makes more sense thank you all so much for your help.

Answer 80

yw

Answer 81

ohh, my pdf has a typo in it. where the integral is. I'll fix that, hold.

Answer 82

okay

Answer 83

there fixed it.

Answer 84

it had a and b instead of z and w, not that it matters, but ....
it was a nice post, and tnx everyone!

Answer 85

thanks.

Answer 86

There is an identity for cos(x) that I do not remember, it complements \[\lim_{x \rightarrow 0} \frac{\sin(x)}{x}=1\] Something like.. \[\lim_{x \rightarrow 0} \frac {1-\cos(x)}{x}=0\] I think..

Answer 87

Anyway, good luck with Calculus!!

Answer 88

i'll try finding it in my book and thank you

Answer 89

http://www.cliffsnotes.com/math/calculus/calculus/limits/limits-involving-trigonometric-functions

Answer 90

yes, correct, Jhanny