Practice questions for exam Q 3. find the limit h goes to 0 of (1-(1+h)^(1/3))/h relating it to a derivative. the solution is suppose to be 1/3 but I get different: = (h goes 0)lim {(-1)[(x^(1/3)+(x+h)^(1/3)]}/h at x=1 is equal to -1(x^(1/3))'at x=1 equal -1*1/3 =-1/3 where I am wrong? I use the factor -1 to transform f(x)-f(x+h)overh in f(x+h)-f(x)overh to get the right term of the derivative.

Question

Practice questions for exam Q 3.  find the limit h goes to 0 of (1-(1+h)^(1/3))/h relating it to a derivative. the solution is suppose to be 1/3 but I get different:  = (h goes 0)lim {(-1)[(x^(1/3)+(x+h)^(1/3)]}/h at x=1 is equal to  -1(x^(1/3))'at x=1 equal -1*1/3 =-1/3  where I am wrong? I use the factor -1 to transform f(x)-f(x+h)overh in f(x+h)-f(x)overh to get the right term of the derivative.

phi · Answer

You are correct.  Who ever wrote up the question solved
$$\lim_{h \rightarrow 0}\frac{ \sqrt[3]{1+h} -\sqrt[3]{1}}{ h }$$
which is -1 times the question asked.   

in other words,  the limit is -1/3  (not 1/3)

anonymous · Answer

thanks Phi!, your answer made me sure.