как се намира границата ?

[Гост]

Здравейте , как се намира
лим.x->a , (sin(x) - sin(a)) / (x-a)

[ammornil]

Здравейте , как се намира
лим.x->a , (sin(x) - sin(a)) / (x-a)

[tex]\sin{x}-\sin{a}=2\sin{\frac{x-a}{2}}\cos{\frac{x+a}{2}}[/tex]

[tex]\lim_{x \to a}{\frac{2\sin{\frac{x-a}{2}}\cos{\frac{x+a}{2}}}{2\frac{x-a}{2}}}=\lim_{x \to a}{\left(\frac{\sin{\frac{x-a}{2}}}{\frac{x-a}{2}}.\cos{\frac{x+a}{2}}\right)}=\lim_{x \to a}{ \frac{\sin{\frac{x-a}{2}}}{\frac{x-a}{2}}}.\lim_{x \to a}{\cos{\frac{x+a}{2}}}=...=1.\cos\frac{a+a}{2}=\cos{2a}[/tex]

[ammornil]

Здравейте , как се намира
лим.x->a , (sin(x) - sin(a)) / (x-a)

[tex]\sin{x}-\sin{a}=2\sin{\frac{x-a}{2}}\cos{\frac{x+a}{2}}[/tex]

[tex]\lim_{x \to a}{\frac{2\sin{\frac{x-a}{2}}\cos{\frac{x+a}{2}}}{2\frac{x-a}{2}}}=\lim_{x \to a}{\left(\frac{\sin{\frac{x-a}{2}}}{\frac{x-a}{2}}.\cos{\frac{x+a}{2}}\right)}=\lim_{x \to a}{ \frac{\sin{\frac{x-a}{2}}}{\frac{x-a}{2}}}.\lim_{x \to a}{\cos{\frac{x+a}{2}}}=...=1.\cos\frac{a+a}{2}=\cos{2a}[/tex]

отговорът е
[tex]\lim_{x \to a}{\frac{2\sin{\frac{x-a}{2}}\cos{\frac{x+a}{2}}}{2\frac{x-a}{2}}}=\lim_{x \to a}{\left(\frac{\sin{\frac{x-a}{2}}}{\frac{x-a}{2}}.\cos{\frac{x+a}{2}}\right)}=\lim_{x \to a}{ \frac{\sin{\frac{x-a}{2}}}{\frac{x-a}{2}}}.\lim_{x \to a}{\cos{\frac{x+a}{2}}}=...=1.\cos\frac{a+a}{2}=\cos{a}[/tex]
, а не [tex]\cos{2a}[/tex]. Забравил съм делението на 2.
[tex][/tex]