G Gatineau New member Joined Apr 18, 2022 Messages 4 Feb 19, 2023 #1 Given f'(5)=-2 and g'(5)=7 Evaluate the limit as h approaches zero of (f(5+6h^2 ) - f(5-2h^2 ))(g(5+h) - g(5-h))/h^3 I know this type of problem is just algebraic manipulation (like most limits), but I really didn't manage to solve it.
Given f'(5)=-2 and g'(5)=7 Evaluate the limit as h approaches zero of (f(5+6h^2 ) - f(5-2h^2 ))(g(5+h) - g(5-h))/h^3 I know this type of problem is just algebraic manipulation (like most limits), but I really didn't manage to solve it.
blamocur Elite Member Joined Oct 30, 2021 Messages 3,223 Feb 19, 2023 #2 I'd start by using formula [imath]f(x + h) = f(x) + f^\prime(h) + O(h^2)[/imath].
stapel Super Moderator Staff member Joined Feb 4, 2004 Messages 16,550 Feb 20, 2023 #3 Gatineau said: Given f'(5)=-2 and g'(5)=7 Evaluate the limit as h approaches zero of (f(5+6h^2 ) - f(5-2h^2 ))(g(5+h) - g(5-h))/h^3 I know this type of problem is just algebraic manipulation (like most limits), but I really didn't manage to solve it. Click to expand... I *think* the limit is as follows: [imath]\displaystyle \lim_{h \rightarrow 0} \frac{\left[f(5+6h^2) - f(5 - 2h^2)\right]\left[g(5+h) - g(5-h)\right]}{h^3}[/imath] This might be split apart as: [imath]\displaystyle \frac{g(5+h)-g(5-h)}{h}[/imath] ...and: [imath]\displaystyle \frac{f(5+6h^2) - f(5-2h^2)}{h^2}[/imath] Eliz.
Gatineau said: Given f'(5)=-2 and g'(5)=7 Evaluate the limit as h approaches zero of (f(5+6h^2 ) - f(5-2h^2 ))(g(5+h) - g(5-h))/h^3 I know this type of problem is just algebraic manipulation (like most limits), but I really didn't manage to solve it. Click to expand... I *think* the limit is as follows: [imath]\displaystyle \lim_{h \rightarrow 0} \frac{\left[f(5+6h^2) - f(5 - 2h^2)\right]\left[g(5+h) - g(5-h)\right]}{h^3}[/imath] This might be split apart as: [imath]\displaystyle \frac{g(5+h)-g(5-h)}{h}[/imath] ...and: [imath]\displaystyle \frac{f(5+6h^2) - f(5-2h^2)}{h^2}[/imath] Eliz.