Question

find the value of lim tan(pi/4 + h)-tan(pi/4)/h as h approaches 0
help :(

Answer 1

\[\lim_{h \rightarrow 0}\frac{\tan(\frac{\pi}{4}+h)-\tan(\frac{\pi}{4})}{h}\]
if you know derivative of trig functions this can be quick
if not then this will be a little long

Answer 2

uhm yeah i know the derivative of trig functions

Answer 3

\[\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}=f'(x) \\ \lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h}=f'(x)|_{x=a}\]

Answer 4

notice your f(x)=tan(x)
and a is?

Answer 5

so just find f'(x)
then plug in the a value you have there

Answer 6

First I need you to use the identity\[\tan (A +B)=\frac{ \tan A +\tan B }{ 1-\tan A \tan B}\]to simplify the limit expression.

Answer 7

guess knowing the derivative of tangent is out of the question here huh?

Answer 8

Yes it is same as fin ding the derivative and evaluating however they want you find the derivative from first principles here so even though @myininaya is correct she is misguiding you unfortunately.

Answer 9

uhm okay i don't quite understand

Answer 10

What part do you not understand??

Answer 11

i know what the correct answer is i just don't know how to get it

Answer 12

I don't believe I'm misguiding anyone.
If they know the derivatives of the trig function and the definition of derivative, then I don't quite see how I'm misguiding.

Answer 13

Do you not understand the identity I told you to use?

Answer 14

no i don't

Answer 15

And as they said they do know the derivative of the trig functions.

Answer 16

uhm so @myininaya can you just help me?

Answer 17

\[\lim_{h \rightarrow 0}\frac{\tan(\frac{\pi}{4}+h)-\tan(\frac{\pi}{4})}{h}=(\tan(x))'|_{x=\frac{\pi}{4}}\]
this was the way I suggested

Answer 18

@dolphins121 I'll show you the first step\[\lim_{h \rightarrow 0}\frac{ \tan (\frac{ \pi }{ 4 }+h)-\tan \frac{ \pi }{ 4 } }{ h }\]\[=\lim_{h \rightarrow 0}\frac{ \frac{ \tan \frac{ \pi }{ 4 }+\tan h }{ 1-\tan \frac{ \pi }{ 4 }\tan h }-\tan \frac{ \pi }{ 4 } }{ h }\]Now can you take over from here.

Answer 19

find the derivative of tan(x)
then evaluate that derivative at x=pi/4

Answer 20

and you are done

Answer 21

wait okay so the derivative of tanx is sec^2x so what do i do from there?

Answer 22

I kept loosing my internet connection and had to start over again. Plus they want it done from first principles and not take the derivative directly. That's where @myininaya is misguiding you!! I am a teacher who signs in here whenever I have free time so that's why I can say that @myininaya you are misguiding him.

Answer 23

I'm not misguiding anyone. I have been a teacher as well. You aren't the only one who teaches.

Answer 24

There are multiple ways to answer a question. Please do not get upset if I come up with another way that is different from yours.

Answer 25

Yes\[\frac{ d }{ dx }\tan x =\sec ^{2}x\]and you can substitute pi/4 in for x but once again you are supposed to arrive at the answer by using the definition (first principles)

Answer 26

And telling someone I'm misguiding them sounds like I'm leading them to a wrong answer and makes it sounds like I'm not even giving them a correct way to follow.

Answer 27

Says who?!

Answer 28

I agree with @myininaya . Her method is shorter and more efficient.

Answer 29

You're missing the point!!! @myininaya I know there are various ways and you are not wrong but he is supposed to use first principles!!! Go ask your own professor!

Answer 30

Again you people are missing the point but whatever you don't have to believe me!! If this is an assignment that will be marked, you'll find out the hard way. Good Luck!!

Answer 31

okay but it doesn't say that i need to use first principles so @myininaya can you keep helping me. i sort of understood i just need some guidance

Answer 32

Do you know the derivative of tan(x) @dolphins121 ?

Answer 33

yes its sec^2x

Answer 34

now just plug in pi/4

Answer 35

\[\lim_{h \rightarrow 0}\frac{\tan(\frac{\pi}{4}+h)-\tan(\frac{\pi}{4})}{h}=(\tan(x))'|_{x=\frac{\pi}{4}} \\ =\sec^2(x)|_{x=\frac{\pi}{4}}=\sec^2(\frac{\pi}{4})= [\sec(\frac{\pi}{4})]^2\]

Answer 36

do you know cos(pi/4)
because sec(pi/4) will the reciprocal of that

Answer 37

yeah it would be 1 divided by sqrt2/2

Answer 38

\[(\frac{1}{\frac{\sqrt{2}}{2}})^2=(\frac{2}{\sqrt{2}})^2\]

Answer 39

then you have 2^2/2

Answer 40

I will let you finish that

Answer 41

okay so then that simplifies to 2

Answer 42

\[\lim_{h \rightarrow 0}\frac{\tan(\frac{\pi}{4}+h)-\tan(\frac{\pi}{4})}{h}=(\tan(x))'|_{x=\frac{\pi}{4}} \text{ by definition of derivative } \\ =\sec^2(x)|_{x=\frac{\pi}{4}}=\sec^2(\frac{\pi}{4})= [\sec(\frac{\pi}{4})]^2 \]
\[=(\frac{2}{\sqrt{2}})^2=\frac{2^2}{2}=\frac{4}{2}=2\]
we can add a little note up there so no points will taken

Answer 43

okay but wouldn't the whole thing be zero because you have sec^2(pi/4+h)-sec^2(pi/4)/h and when you work everything out you'll be subtracting 2-2? and since h is approaching zero the denominator is zero anyways?

Answer 44

are you familiar with this:
\[\lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h}=f'(a)\]

Answer 45

this is the definition of derivative

Answer 46

yes

Answer 47

we had f(x)=tan(x) and a=pi/4

Answer 48

\[f(\frac{\pi}{4}+h)=\tan(\frac{\pi}{4}+h) \\ f(\frac{\pi}{4})=\tan(\frac{\pi}{4}) \\\]

Answer 49

\[\lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h}=f'(a) \\ \lim_{h \rightarrow 0} \frac{\tan(\frac{\pi}{4}+h)-\tan(\frac{\pi}{4})}{h}=(\tan(a))'|_{a=\frac{\pi}{4}}=\sec^2(a)|_{x=\frac{\pi}{4}}=\sec^2(\frac{\pi}{4})=2\]
do you that our limit expression is in the same form
where f(x)=tan(x) and a=pi/4

Answer 50

uhm yeah i get that. but don't we plug in zero for h? that's what's confusing me

Answer 51

we don't have any h's anymore when we get to the derivative of tan(x) part

Answer 52

oh. okay so can you just show me how it would look to see if i understand

Answer 53

But I already showed you
\[\lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h}=[f(x)]'|_{x=a}\ \text{ we have } f(x)=\tan(x) \text{ and } a=\frac{\pi}{4} \\ \lim_{h \rightarrow 0}\frac{\tan(\frac{\pi}{4}-h)-\tan(\frac{\pi}{4})}{h}=[\tan(x)]'|_{x=\frac{\pi}{4}} \text{ note: no h's here }\]

Answer 54

say we have this limit expression
\[\lim_{h \rightarrow 0}\frac{\sqrt{4+h}-\sqrt{4}}{h}=[\sqrt{x}]'_{x=4}=\frac{1}{2\sqrt{x}}|_{x=4}=\frac{1}{2 \sqrt{4}}=\frac{1}{2(2)}=\frac{1}{4}\]

Answer 55

or you could rationalize the numerator and still wind up with 1/4
\[\lim_{h \rightarrow 0}\frac{4+h-4}{h(\sqrt{4+h}+\sqrt{4})}=\lim_{h \rightarrow 0}\frac{1}{\sqrt{4+h}+\sqrt{4}}=\frac{1}{\sqrt{4+0}+\sqrt{4}}=\frac{1}{2+2}=\frac{1}{4}\]

Answer 56

still makes no sense?

Answer 57

oh okay. i get it. thanks

Answer 58

here is another example just in case
\[\lim_{h \rightarrow 0}\frac{\cos(\frac{\pi}{4}+h)-\cos(\frac{\pi}{4})}{h}=(\cos(x))'_{x=\frac{\pi}{4}}=-\sin(x)|_{x=\frac{\pi}{4}}=-\sin(\frac{\pi}{4})=- \frac{\sqrt{2}}{2}\]

Answer 59

so whenever you have
\[\lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h} \text{ you can skip to } (f(x))'_{x=a}\]

Answer 60

oh okay. well thanks

Answer 61

Well that was fun ;)