lim x tends to 0 [1/cotx -1/1+x2]
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actually it should be cos x
The limit `lim_(x->0) (1-cosx)/x^2` has to be determined.
If we substitute x = 0, the result `(1-cosx)/x^2` is the indeterminate form `0/0` . In this case it is possible to use l'Hospital's rule and replace the numerator and denominator with their derivatives.
`(1 - cos x)' = sin x`
`(x^2)' = 2x`
The limit is now:
`lim_(x->0) sin x/(2x)`
If the substitution x = 0 is made here, the result is again the indeterminate form `0/0` . Using l'Hospital's rule, replace the numerator and denominator with their derivatives. This gives:
`lim_(x->0) cos x/2`
Substituting x = 0 gives the result 1/2.
The limit `lim_(x->0) (1-cosx)/x^2 = 1/2` .
The limit `lim_(x->0) (1-cosx)/x^2` has to be determined.
If we substitute x = 0, the result `(1-cosx)/x^2` is the indeterminate form `0/0` . In this case it is possible to use l'Hospital's rule and replace the numerator and denominator with their derivatives.
`(1 - cos x)' = sin x`
`(x^2)' = 2x`
The limit is now:
`lim_(x->0) sin x/(2x)`
If the substitution x = 0 is made here, the result is again the indeterminate form `0/0` . Using l'Hospital's rule, replace the numerator and denominator with their derivatives. This gives:
`lim_(x->0) cos x/2`
Substituting x = 0 gives the result 1/2.
The limit `lim_(x->0) (1-cosx)/x^2 = 1/2` .
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