Rewriting a limit as a derivative

[NavalMonte]

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]

What function do you think you'd be taking the derivative of, do you think?

 

[NavalMonte]

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]

Well, when you calculate a derivative, it's the derivative of a function. What's the only function in sight here?

 

[soroban]

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:

. . $$\text{Given: }\:f(x) \,=\,\sec x$$
. . $$\text{Find }f'(x)\text{ at }x = \pi\text{ by definition.}$$$$\frac{f(\pi+h) - f(x)}{h} \;=\;\frac{\sec(\pi+h) - \sec(\pi)}{h}$$

. . . $$=\;\frac{\sec(\pi+h) - (-1)}{h} \;=\;\frac{\sec(\pi+h) + 1}{h}$$

Note that: .$$\sec(\pi+h) \:=\:\frac{1}{\cos(\pi+h)} \:=\:\frac{1}{-\cos h}$$

We have: .$$\frac{-\frac{1}{\cos h} +1}{h} \:=\:\frac{-1 + \cos h}{h\cos h} \:=\:-\frac{1-\cos h}{h\cos h}$$

Multiply by $$\frac{1+\cos h}{1+\cos h}\!:\;-\frac{1-\cos h}{h\cos h}\cdot\frac{1+\cos h}{1+\cos h}$$

. . . $$=\;-\frac{1-\cos^2h}{h\cos h(1+\cos h)} \;=\;-\frac{\sin^2h}{h\cos h(1+\cos h)}$$

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

 

[Deveno]

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]

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]

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]

Meant to say Sec(x).

Silly mistake (Doh)