14a) Let eʳ=2ˣ; r=xln(2); so the limit expression becomes (eˣˡⁿ⁽²⁾-1)/x=(1+xln(2)+x²ln²(2)/2!+... -1)/x=ln(2)+xln²(2)/2!+... (using the Taylor expansion for eʳ). Therefore the limit as x➝0 is ln(2)=0.6931 approx.
14b) log₁₀e=1/ln(10)=0.4345 approx.
Method: x=10^log₁₀(x); and 10=e^ln(10); so x=(e^ln(10))^log₁₀(x)=e^(ln(10)log₁₀(x)); taking logs of each side:
ln(x)=ln(10)log₁₀(x) and log₁₀(x)=ln(x)/ln(10).
ln(x)=ln(1+(x-1))=(x-1)-(x-1)²/2+... (Taylor expansion)
Divide by x-1 and we get 1-(x-1)/2+... When x➝1, ln(x)➝1 so log₁₀(x)➝1/ln(10)=0.4345 approx.
15a) ln(3)=1.0986 approx. See 14a) for explanation, substituting 3 for 2.
15b) 1 (see 14b) method using Taylor expansion for ln(x))
16a) 0.25 Method: Let x=5+h where h is small. The expression becomes (√(5+h-1)-2)/h.
So (√(4+h)-2)/h=(2√(1+h/4)-2)/h=(2(1+h/8+...)-2)/h=(2+h/4+...-2)/h=1/4=0.25 when h is very small.
b) 0.6: sin(3x)≈3x when x is small so sin(3x)/5x≈3x/5x=3/5=0.6 as x➝0.
17a) 0.125 (see 16a) for method using h)
b) 3.5 (see 16b) for method)