Evaluate :- lim x → 0 (1-cosx root(cos2x))/(e^(x^2)-1)

Question

Vikas TU · Answer

Apply L-Hospital. it would become = > (0 + sinx root(cos2x) - cosx * 0.5*1/root(cos2x) * sin2x * 2)/e^(x^2) * 2x apply again . means again apply L-Hospital. if 0/0 indeterminat forms.

Divyansh Thada · Answer

please apply brackets appropriately wherever necessary please apply brackets appropriately wherever necessary

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Evaluate :- lim x → 0 (1-cosx root(cos2x))/(e^(x^2)-1)

Evaluate :-

lim_{x → 0} (1-cosx root(cos2x))/(e^(x^2)-1)

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Evaluate :- lim x → 0 (1-cosx root(cos2x))/(e^(x^2)-1)

Evaluate :- lim x → 0 (1-cosx root(cos2x))/(e^(x^2)-1)

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Evaluate :-

lim_{x → 0} (1-cosx root(cos2x))/(e^(x^2)-1)