Question

How do you find the limit `lim_(h->0)(sqrt(1+h)-1)/h` ?

Answer 1

`\frac{1}{2}`

Explanation:

The limit presents an undefined form `0/0`. In this case, you may use de l'hospital theorem, that states

`lim \frac {f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)}`

The derivative of the numerator is

`\frac{1}{2sqrt(1+h)}`

While the derivative of the denominator is simply `1`.

So,

`\lim_{x\to 0} \frac{f'(x)}{g'(x)} = \lim_{x\to 0} \frac{\frac{1}{2sqrt(1+h)}}{1} =\lim_{x\to 0} \frac{1}{2sqrt(1+h)}`

And thus simply

`\frac{1}{2sqrt(1)}=\frac{1}{2}`

Answer 2

`= 1/2`

Explanation:

If you are unaware of l'hopitals rule...

Use:

`(1+x)^n = 1 + nx + (n(n-1))/(2!) x^2 + ...`

`=> (1 + h)^(1/2) = 1 + 1/2h - 1/8 h^2 + ...`

`=> lim_( h to 0) ((1 + 1/2 h - 1/8h^2 + ...) - 1 )/ h`

`=> lim_( h to 0) ( 1/2 h - 1/8h^2 + ... )/ h`

`=> lim_( h to 0) ( 1/2 - 1/8 h + ... )`

`= 1/2`