Find f(x + Ax) by substituting x + Ax in place of x in f(x) = 9x² – 5x.2f(x + Ax) = 9( x+ Ax)² – 5( x + Axx + Arx +AxStep 2Simplify the above equation.f(x + Ax) = 9x² + 18 xAx+ 9Ax2 – 5x –Axx· ArStep 3Now find the value of f(x + Ax) – f(x).f(x + Ax) – f(x) = (9x2 + 18xAx + 9Ax2 – 5x – 5Ax) – (9x2 – 5x)9x2 + 18xAx + 9Ax² – 5x – 5Ax – 9x² + 5x18xAx + 9Ax2 – 5Ax|18r · Ar + 9. Ax² – 5. AxStep 4The limit is rewritten as follows.f(x + Дх) — f(x)18хДх + 9Дх2 — 5ДхlimAx → 0limAx → 0AxAxBecause the limit cannot be evaluated by direct substitution, divide out the common factor Ax and simplify.f(x + Дх) — f(x)Ax()limAx → 0limAx → 0AxAxlimAx → 05

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f(x + Ax) = 9( x+ Ax p² – 5( x + Ax %3D x +Ax x +Ar ep 2 Simplify the above equation. f(x + Ax) = 9x² + 18 xAx + 9Ax2 – 5x – x· Ar ep 3 Now find the value of f(x + Ax) – f(x). f(x + Ax) – f(x) = (9x2 + 18xAx + 9Ax² – 5x – 5Ax) – (9x² – 5x) - 9x2 + 18хДх + 9Дх2 - 5х — 5Дх — 9х2 + 5х %3D 18xAx + 9Ax² – 5Ax| 18ε Δ+9. Δ5. Δε ep 4 The limit is rewritten as follows. f(x + Дх) — f(x) 18×AX + 9Ax2 – 5Ax lim Ax → 0 lim Ax → 0 %3D Ax Ax Because the limit cannot be evaluated by direct substitution, divide out the common factor Ax and sim Ax( lim Ax → 0 f(x + Ax) – f(x) lim Ax → 0 Ax Ax