Step by step solution to "What do the limh→0sin⁡hh and limh→0cosh−1h have to..."

amanf

Answered question

2021-10-14

What do the $lim_{h \to 0} \frac{\sin h}{h}$ and $lim_{h \to 0} \frac{\cosh - 1}{h}$ have to do with the derivatives of the sine and cosine functions? What are the derivatives of these functions?

Answer & Explanation

Ezra Herbert

Skilled2021-10-15Added 99 answers

Step 1
Explain the limit of $lim_{h \to 0} \frac{\sin h}{h}$
Step 2
Here, $lim_{h \to 0} \frac{\sin h}{h}$ is the derivative of $\sin x$ at $x = 0$ .
$f^{'} (0) = lim_{h \to 0} \frac{f (0 + h) - f (0)}{h}$
$= lim_{h \to 0} \frac{\sin h - \sin 0}{h}$
$= lim_{h \to 0} \frac{\sin h}{h}$
$= lim_{h \to 0} \frac{\cos h}{1}$ $[∴ lim_{h \to 0} \frac{\sin}{h} = \frac{0}{0}]$
$= \cos 0$
$= 1$

Jeffrey Jordon

Expert2021-10-19Added 2605 answers

Explain the limit of $lim_{h \to 0} \frac{\cos h - 1}{h}$

Here, $lim_{h \to 0} \frac{\cos h - 1}{h}$ is the derivative of $\cos x$ at $x = 0$

$f^{'} (0) = lim_{h \to 0} \frac{f (0 + h) - f (0)}{h}$

$= lim_{h \to 0} \frac{\cos h - \cos 0}{h}$

$= lim_{h \to 0} \frac{- \cos h - 1}{h}$

$= lim_{h \to 0} \frac{- \sin h}{1}$ $[∴ lim \frac{\cos h - 1}{h} = \frac{0}{0}]$

$= - \sin 0$

$= 0$

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