Question

[(msinx – ncosx)/(x – α)] for lim x→α = ? [Where m sin α – n cos α = 0, m, n ∈ N, π < α < (3π/2)]

(a) √(m² + n²)

(b) √(m² – n²)

(c) – √(m² + n²)

(d) – [1/√(m² – n²)]

Answer 1

The correct option (a) √(m² + n²)

Explanation:

given:

lim_x→α [(m sin x – n cos x)/(x – α)]

= m ∙ lim_x→α [{sin x – (n/m)cos x}/(x – α)]

= m ∙ lim_x→α [(sinx – tanα ∙ cosx)/(x – α)] ---- msinα = n cosα

= m ∙ lim_x→α [(cos x + tan α ∙ sin x)/1]

= m ∙ lim_x→α cos x + [(sin α)/(cos α)] ∙ sin x

= m ∙ [cos x + {(sin² α)/(cos α)}]

= m ∙ [1/(cos α)] = [n/(sin α)]  (1)

Consider  m² + n² = m² + m² [(sin² α)/(cos² α)]

= m² [(cos² α + sin² α)/(cos² α)]

= [m²/(cos² α)]

∴  √(m² + n²) = [m/(cos α)]

∴ from (1) value of given limit is √(m² + n²)