Given f(x)=1+x. The value of limh→0h2f(3+2h2)−f(3−3h2) is....
Explanation
First find the derivative of the function
f(x)=1+x
f′(x)=21+x1
If y=f(g(x)) then the first derivative is y′=g′(x)⋅f′(g(x)). Check the limit value
h→0limh2f(3+2h2)−f(3−3h2)=00
Then we use another method, L'Hospital's rule or the first derivative
h→0limh2f(3+2h2)−f(3−3h2)
=h→0lim2h4hf′(3+2h2)−(−6h)f′(3−3h2)
=h→0lim24f′(3+2h2)+6f′(3−3h2)
=h→0lim210f′(3+2h2)
=h→0lim5f′(3+2h2)
=5f′(3+2⋅02)
=5f′(3)
=5(21+31)
=45