Question

Evaluate the Given limit: \(\lim_{x\rightarrow 0}\) sinax =\(\frac{bx}{ax}\)+ sinbx a,b,a+b ≠ 0

Answer 1

Given:

lim_x→0 [ sin(ax) / (bx) + sin(bx) / (ax) ]

where a ≠ 0, b ≠ 0 and a + b ≠ 0

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Step 1: Split the limit

lim_x→0 [ sin(ax)/(bx) + sin(bx)/(ax) ]

= lim_x→0 sin(ax)/(bx) + lim_x→0 sin(bx)/(ax)

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Step 2: Simplify each term

sin(ax)/(bx) = (a/b) · [ sin(ax)/(ax) ]

sin(bx)/(ax) = (b/a) · [ sin(bx)/(bx) ]

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Step 3: Apply standard limits

lim_θ→0 sinθ/θ = 1

So,

lim_x→0 sin(ax)/(ax) = 1
lim_x→0 sin(bx)/(bx) = 1

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Step 4: Evaluate the limit

lim_x→0 [ sin(ax)/(bx) + sin(bx)/(ax) ]

= (a/b) + (b/a)

= (a² + b²) / ab

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Final Answer:

lim_x→0 [ sin(ax)/(bx) + sin(bx)/(ax) ] = (a² + b²) / ab