Any "mathematician" on brainly?????
Check for the existence of limits
[tex] \lim_{n \to0} \frac{ \sqrt{1-cos2x} }{x}
[/tex]
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yes the limit exists
lim_(x->0) (1-cos(2 x))/x
Applying l'Hôpital's rule, we get that
lim_(x->0) (1-cos(2 x))/x | = | lim_(x->0) ( d/( dx)(1-cos(2 x)))/(( dx)/( dx))
| = | lim_(x->0) (2 sin(2 x))/1
| = | lim_(x->0) 2 sin(2 x)
lim_(x->0) 2 sin(2 x)
lim_(x->0) 2 sin(2 x) = 2 sin(2 0) = 0:
Answer: |
| 0
lim_(x->0) (1-cos(2 x))/x
Applying l'Hôpital's rule, we get that
lim_(x->0) (1-cos(2 x))/x | = | lim_(x->0) ( d/( dx)(1-cos(2 x)))/(( dx)/( dx))
| = | lim_(x->0) (2 sin(2 x))/1
| = | lim_(x->0) 2 sin(2 x)
lim_(x->0) 2 sin(2 x)
lim_(x->0) 2 sin(2 x) = 2 sin(2 0) = 0:
Answer: |
| 0
Answered by
0
lim_(x->0) (1-cos(2 x))/x
Applying l'Hôpital's rule, we get that
lim_(x->0) (1-cos(2 x))/x | = | lim_(x->0) ( d/( dx)(1-cos(2 x)))/(( dx)/( dx))
| = | lim_(x->0) (2 sin(2 x))/1
| = | lim_(x->0) 2 sin(2 x)
lim_(x->0) 2 sin(2 x)
lim_(x->0) 2 sin(2 x) = 2 sin(2 0) = 0
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