Calculus Examples

$lim_{h \to 0} \frac{\sin (\frac{π}{4} + h) - \frac{\sqrt{2}}{2}}{h}$

Step 1

Reduce.

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Step 1.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$lim_{h \to 0} \frac{\sin (\frac{π}{4} + h) - \frac{2^{\frac{1}{2}}}{2}}{h}$

Step 1.2

Factor $2$ out of $2^{\frac{1}{2}}$ .

$lim_{h \to 0} \frac{\sin (\frac{π}{4} + h) - \frac{2 \cdot 2^{- \frac{1}{2}}}{2}}{h}$

Step 1.3

Cancel the common factors.

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Step 1.3.1

Factor $2$ out of $2$ .

$lim_{h \to 0} \frac{\sin (\frac{π}{4} + h) - \frac{2 \cdot 2^{- \frac{1}{2}}}{2 (1)}}{h}$

Step 1.3.2

Cancel the common factor.

$lim_{h \to 0} \frac{\sin (\frac{π}{4} + h) - \frac{2 \cdot 2^{- \frac{1}{2}}}{2 \cdot 1}}{h}$

Step 1.3.3

Rewrite the expression.

$lim_{h \to 0} \frac{\sin (\frac{π}{4} + h) - \frac{2^{- \frac{1}{2}}}{1}}{h}$

Step 1.3.4

Divide $2^{- \frac{1}{2}}$ by $1$ .

$lim_{h \to 0} \frac{\sin (\frac{π}{4} + h) - 2^{- \frac{1}{2}}}{h}$

Step 1.4

Move $2^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$lim_{h \to 0} \frac{\sin (\frac{π}{4} + h) - \frac{1}{2^{\frac{1}{2}}}}{h}$

Step 2

Since the function approaches $- \infty$ from the left and $\infty$ from the right, the limit does not exist.

$Does not exist$