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Limits And Derivatives

Multiple Choice Questions

$\lim_{x \to 0} \frac{1 - \cos^{3} (4 x)}{x^{2}} = ?$

$If f (x) = \sin (\cos (x)), then f' (x) = ?$

$\cos (\cos (x))$
$\sin (- \sin (x))$
$(\sin (x)) \cos (\cos (x))$
$(- \sin (x)) \cos (\cos (x))$

$What is the derivative of the function f (x) = e^{\tan (x)} + \ln (\sec (x)) - e^{\ln (x)} at x = \frac{π}{4} ?$

$\frac{e}{2}$
e
2e
4e

$What is \lim_{x \to 0} \frac{\tan (x)}{\sin (2 x)} = ?$

$\frac{1}{2}$
1
2
Limit does not exist

5.
$What is \lim_{h \to 0} \frac{\sqrt{2 x + 3 h} - \sqrt{2 x}}{2 h} = ?$

$\frac{1}{2 \sqrt{2} x}$

$\frac{3}{\sqrt{2} x}$

$\frac{3}{2 \sqrt{2} x}$

$\frac{3}{2 \sqrt{2} x}$

$\frac{3}{2 \sqrt{2} x}$

$\lim_{h \to 0} \frac{\sqrt{2 x + 3 h} - \sqrt{2 x}}{2 h} = \lim_{h \to 0} \frac{\sqrt{2 x + 3 h} - \sqrt{2 x}}{2 h} \times \frac{\sqrt{2 x + 3 h} + \sqrt{2 x}}{\sqrt{2 x + 3 h} + \sqrt{2 x}} = \lim_{h \to 0} \frac{{(\sqrt{2 x + 3 h})}^{2} - {(\sqrt{2 x})}^{2}}{2 h (\sqrt{2 x + 3 h} + \sqrt{2 x})} = \lim_{h \to 0} \frac{2 x + 3 h - 2 x}{2 h (\sqrt{2 x + 3 h} + \sqrt{2 x})} = \lim_{h \to 0} \frac{3 h}{2 h (\sqrt{2 x + 3 h} + \sqrt{2 x})} = \lim_{h \to 0} \frac{3}{2 (\sqrt{2 x + 3 h} + \sqrt{2 x})} = \frac{3}{2 \times 2 \sqrt{2} x} = \frac{3}{4 \sqrt{2} x}$