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#### NavalMonte

##### New member

- Feb 18, 2014

- 7

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

Any hints on how to start?

- Thread starter NavalMonte
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- Thread starter
- #1

- Feb 18, 2014

- 7

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

Any hints on how to start?

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- #2

- Jan 26, 2012

- 4,198

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

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- #3

- Feb 18, 2014

- 7

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

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- #4

- Jan 26, 2012

- 4,198

It would have been considerate to give us the original problem.

I'm forced to

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]

- Feb 15, 2012

- 1,967

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

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

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- #7

- Feb 18, 2014

- 7

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

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.

- Feb 15, 2012

- 1,967

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").

- Thread starter
- #9

- Feb 18, 2014

- 7

Meant to say Sec(x).

Silly mistake

Silly mistake