Question

Can ANYONE HELP ME WITH LIMITS CALCULUS? (Qn 32)

Answer 1

I assume you don't know about derivatives yet. The reason I bring that up is because the given limit is equivalent to the derivative of a certain function and you might recognize it later on.

To compute the limit, in the numerator we can combine the fractions into one:

`\frac1{(x+h)^2}-\frac1{x^2}=(x^2-(x+h)^2)/(x^2(x+h)^2)=(x^2-(x^2+2xh+h^2))/(x^2(x+h)^2)=-((2x+h)h)/(x^2(x+h)^2)`

We're taking the limit as
`h\to0`, which means we're considering
`h` near 0, but not
`h=0`, so that we can simplify:

`\frac{-((2x+h)h)/(x^2(x+h)^2)}h=-(2x+h)/(x^2(x+h)^2)`

Then as
`h\to0`, we have
`2x+h\to2x` and
`x+h\to x`, so that

`\displaystyle\lim_(h\to0)-(2x+h)/(x^2(x+h)^2)=-(2x)/(x^4)=-\frac2{x^3}`

provided that
`x\\eq0`.

(In fact, the limit is equivalent to the derivative of the function
`f(x)=\frac1{x^2}`, whose derivative is
`f'(x)=-\frac2{x^3}`)