Question

[(1/(h+2)^2)-(1/x^2)]/h, and i'm looking for limit as h --> 0.

Answer 1

in you question you have an h+2 when it should be h+x so i will solve with h+x prolly just a typo

\[\lim_{h \rightarrow 0} [ 1/(h+x)^2 - 1/x^2] \times1/h\]
\[=\lim_{h \rightarrow 0} [ 1/(h^2 +2hx+x^2) - 1/x^2] \times1/h\]
\[=\lim_{h \rightarrow 0} [ (x^2 -(h^2+2hx +x^2))/x^2(h^2 +2hx+x^2)] \times1/h\]\[=\lim_{h \rightarrow 0} [ (x^2 -h^2-2hx -x^2)/(h^2x^2 +2hx^3+x^4)] \times1/h\]
\[=\lim_{h \rightarrow 0} [ (-h^2-2hx)/(h^2x^2 +2hx^3+x^4)] \times1/h\]
\[=\lim_{h \rightarrow 0} [ (-h-2x)/(h^2x^2 +2hx^3+x^4)]\]
applying the limit gives
\[ (-(0)-2x)/((0)^2x^2 +2(0)x^3+x^4)\]
\[ =-2x/x^4\]
\[ =-2/x^3\]

Answer 2

so, in line 3 - you multiply it by x^2 to get a common denominator, and then by the top because whatever you do to the bottom, you have to do to the top and vice versa? Is that the reasoning? Thank you.

Answer 3

yea you get common denom so you multiply thetop and bottom left fraction by the denom in the second then multiply top and bottom of teh second fraction by denom of first