Question

Lim x->0 tan^-1x/x solve?

Answer 1

`1`

Explanation:

`lim_(x->0) tan^(-1)x/x->0/0`

So we can try L'Hopital's rule:

For the numerator:

`d/dx tan^-1x=1/(1+x^2)`

and the denominator:

`d/dx(x) = 1`

There fore we get:

`lim_(x->0) tan^(-1)x/x=lim_(x-> 0 )(1/(1+x^2))/1=(1/1)/1=1`

Answer 2

Alternate way of thinking of it, but `underline("less rigorous")`

`lim_(x to 0) ( tan^(-1)(x) ) /x = 1`

Explanation:

One thing to consider is approximations of trigonometric functions when `x` is particulaly small

The first thing to note is that `color(red)(tanx approx x` for `x` being small

This is due to `cosx approx 1` and `sinx = x` for `x` being small

Hence `tanx = sinx / cosx approx x / 1 approx x`

So hence `tan^(-1)x` is simply `tanx` reflected in the line `y = x`

But we know the inverse of `y = x` is just `y = x`

`=> tan^(-1)(x) approx x` for small `x`

`=> tan^(-1)x / x approx x/x approx 1` when `x` is very small

`lim_(x to o) (tan^(-1)x )/ x = lim_(x to 0 ) 1`

`= 1`

We can say this as `x` gets small, to be particular, `x to 0` but we `color(blue)(underline ("couldn't")` do this if we had the limit of something like `x to 1/10`

Like I said at the beggining this method is a very non- rigorous approach, and doesnt always work on all functions

I'd advise you to use `L'H` rule for this problem, as answered prior