Welcome to our community

Be a part of something great, join today!

[SOLVED] Rewriting a limit as a derivative

NavalMonte

New member
Feb 18, 2014
7
I was asked to rewrite the limit as a derivative:

$$\lim_{h\rightarrow 0} \dfrac{sec(\pi + h) + 1}{h}$$

Any hints on how to start?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
What function do you think you'd be taking the derivative of, do you think?
 

NavalMonte

New member
Feb 18, 2014
7
What function do you think you'd be taking the derivative of, do you think?
I'm sorry but I'm not sure what you're asking.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
Well, when you calculate a derivative, it's the derivative of a function. What's the only function in sight here?
 

soroban

Well-known member
Feb 2, 2012
409
Hello, NavalMonte!

It would have been considerate to give us the original problem.
I'm forced to guess what you are asking.


I was asked to rewrite the limit as a derivative:
$$\lim_{h\to0} \dfrac{\sec(\pi + h) + 1}{h}$$

I wager that the problem said:

. . [tex]\text{Given: }\:f(x) \,=\,\sec x[/tex]
. . [tex]\text{Find }f'(x)\text{ at }x = \pi\text{ by definition.}[/tex]


[tex]\frac{f(\pi+h) - f(x)}{h} \;=\;\frac{\sec(\pi+h) - \sec(\pi)}{h}[/tex]

. . . [tex]=\;\frac{\sec(\pi+h) - (-1)}{h} \;=\;\frac{\sec(\pi+h) + 1}{h}[/tex]

Note that: .[tex]\sec(\pi+h) \:=\:\frac{1}{\cos(\pi+h)} \:=\:\frac{1}{-\cos h}[/tex]

We have: .[tex]\frac{-\frac{1}{\cos h} +1}{h} \:=\:\frac{-1 + \cos h}{h\cos h} \:=\:-\frac{1-\cos h}{h\cos h}[/tex]

Multiply by [tex]\frac{1+\cos h}{1+\cos h}\!:\;-\frac{1-\cos h}{h\cos h}\cdot\frac{1+\cos h}{1+\cos h}[/tex]

. . . [tex]=\;-\frac{1-\cos^2h}{h\cos h(1+\cos h)} \;=\;-\frac{\sin^2h}{h\cos h(1+\cos h)} [/tex]

. . . [tex]=\;-\frac{\sin h}{h}\cdot\frac{\sin h}{\cos h(1+\cos h)} [/tex]


Thus: .[tex]\lim_{h\to0}\left[-\frac{\sin h}{h}\cdot\frac{\sin h}{\cos h(1 + \cos h)} \right] \:=\:-1\cdot\frac{0}{1(2)} \:=\:0[/tex]
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
What Ackbach was getting at is:

What is $\sec(\pi) = \dfrac{1}{\cos(\pi)}$?

Do you see an expression in your formula that is the negative of this?
 

NavalMonte

New member
Feb 18, 2014
7
Rewriting this limit into a derivative:

$$\lim_{h\rightarrow 0} \dfrac{sec(\pi + h) + 1}{h}$$

Looks like it came from:

$$\lim_{h\rightarrow 0} \dfrac{f(a + h) - f(a)}{h}$$

I have to ask:

a=?
f(a+h)=?

So,
a = $\pi$
f(a + h) = sec($\pi$ + h)

therefore,
f(a) = Sec($\pi$) = -1

Plugging it back in works out:


$$\lim_{h\rightarrow 0} \dfrac{sec(\pi + h) - (-1)}{h} => \lim_{h\rightarrow 0} \dfrac{sec(\pi + h) + 1}{h}$$

So the answer would be:

$$\lim_{h\rightarrow 0} \dfrac{sec(\pi + h) + 1}{h} = \dfrac{d}{dx}[cos(x)]$$

Would this be correct?
I had a professor give me a hint, but I'm not to sure if this would be what the question was asking for.
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Nope.

You had everything right until this point:

$\displaystyle \lim_{h \to 0} \frac{\sec(\pi + h) + 1}{h}$

now if $\sec(\pi) = -1$, we can re-write THAT as:

$\displaystyle \lim_{h \to 0} \frac{\sec(\pi + h) - \sec(\pi)}{h}$

which looks like it's the derivative of what we're calling $f$, which is...?

(Hint: it's NOT "cosine").
 

NavalMonte

New member
Feb 18, 2014
7
Meant to say Sec(x).

Silly mistake (Doh)