Question

Question #f1b4f

Answer 1

`lim_(x->0) sin(x^2)/x = 0`

Explanation:

First, multiply the top and bottom of the limit by `x`:

`lim_(x->0) (x*sin(x^2))/x^2`

Now create a new variable `u = x^2`.
Note that `x = sqrtu`, and as `x->0`, `u->0`, since `(0)^2 = 0`.
We can therefore rewrite the limit as:

`lim_(u->0) sqrtu * (sinu/u)`

`= lim_(u->0)sqrtu * lim_(u->0)sinu/u`

The second limit is one of the famous calculus limits `lim_(x->0)sinx/x = 1`.

`= lim_(u->0)sqrtu * (1)`

`= 0 * 1`

`= 0`

Final Answer

Answer 2

Use L'hopital's rule to find that `lim_(x->0)sin(x^2)/x=0`

Explanation:

For `x=o, sin(x^2)/x=0/0` so we can use L'hopital's rule:

`lim_(x->a) (f(x))/(g(x))=lim_(x→a)(f'(a))/(g'(a))` given that `limf(x)` and `limg(x)` exist

`lim_(x→0) sin(x^2)/x = lim_(x→0) ((sin(x^2))')/(x') = lim_(x→0)2xcos(x^2)`

Letting `x` equal zero gives `lim_(x→0) sin(x^2)/x = 0`

Answer 3

`0`

Explanation:

formula`rArr lim_(x->0) sinx/x = 1`

`rArr lim_(x->0) sinx^2/x`

Now, multiply and divide x
`rArr lim_(x->0) (sinx^2 * x)/(x * x)`
`rArr lim_(x->0) (sinx^2 * x)/x^2`
`rArr lim_(x->0) 1 * x`

`rArr 0`

ENJOY MATHS!!!!!