Use tangent line approximation to derive an estimate for (1 + x)n , when x is near 0, and n is any real number.
2 answers
Just wanted to make one more clarification, the equation is supposed to be (1 + x)^n.
y = (1+x)^n when x is near zero y is 1
dy/dx = slope = n (1+x)^(n-1)
when x is near 0, slope is near n
so
y = n x + 1
dy/dx = slope = n (1+x)^(n-1)
when x is near 0, slope is near n
so
y = n x + 1
1 answer