Question

The function f(x) = tanx for x = 0; f(0) = 1 at x = 0

Answer 1

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fx = tanx

${lim} \atop \: {h - 0}$

$\frac{f(x + h) - f \: (x)}{h}$

f (X + h) = tan (X+h)

${lim} \atop \: {h - 0} \:$

$\frac{ \tan(x + h) - \tan \: x ) }{0}$

${lim} \atop \: {h - 0}$

$\frac{ \frac{ \tan \: x + \tan \: h }{1 - \tan \: x \: \tanh } - \tan \: x \frac{(1 - \tan \: x \: \tan \: h }{1 - \tan \: x \: \tan \: h } }{h}$

${lim} \atop \: {h - 0}$

$\frac{ \tan \: x \: + \tan \: h - \tan \: x \: + { \tan \: }^{2} x \tan \: h }{h \: (1 - \tan \: x \: \tan \: h) }$

${lim} \atop \: {h - 0}$

$\frac{ \tan \: (1 - { \tan }^{2} x) }{h(1 - \tan \: x \: \tan \: h) }$

${lim} \atop \: {h - 0}$

$\frac{ \sin \: h }{h} \frac{ { \sec }^{2}x }{ \cos \: h(1 - \tan \: x \: \tan \: h) }$

${lim} \atop \: {h - 0}$

$\frac{ \sin \: h }{h}$

${lim} \atop \: {h - 0}$

$\frac{ { \sec}^{2} x }{ \cos \: h(1 - \tan \: x \: \tan \: h)}$

$1 \times \frac{ { \sec}^{2} x}{1 \times (1 - 0)}$

$fx \: = { \sec }^{2} x$

f(0) = sec²ⁿ

f(0) = 1

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