Question

How to do this?

Answer 1

`lim_(x->0) (sin 4x)/(tan 5x) = 4/5`

Explanation:

Quick method

Note that:

`lim_(t->0) sin t / t = 1`

`lim_(t->0) tan t / t = 1`

That is, both `sin t` and `tan t` behave like `t` for small values of `t`.

So:

`lim_(x->0) (sin 4x) / (tan 5x) = lim_(x->0) (4x)/(5x) = 4/5`

L'Hôpital's rule method

Note that:

`lim_(x->0) sin 4x = 0`

`lim_(x->0) tan 5x = 0`

So the requested limit is an indeterminate form `0/0` and hence L'Hôpital's rule applies.

`lim_(x->0) (sin 4x)/(tan 5x) = lim_(x->0) (d/(dx) sin 4x)/(d/(dx) tan 5x)`

`color(white)(lim_(x->0) (sin 4x)/(tan 5x)) = lim_(x->0) ((d/(dx) 4x)(cos 4x))/((d/(dx) 5x)(sec^2 5x))`

`color(white)(lim_(x->0) (sin 4x)/(tan 5x)) = lim_(x->0) (4 cos 4x)/(5 sec^2 5x)`

`color(white)(lim_(x->0) (sin 4x)/(tan 5x)) = (4 * 1)/(5 * 1)`

`color(white)(lim_(x->0) (sin 4x)/(tan 5x)) = 4/5`

Answer 2

`lim_(x->0)sin(4x)/tan(5x)=4/5`

Explanation:

`lim_(x->0)sin(4x)/tan(5x)`

Since `tan(theta)=sin(theta)/cos(theta)`, we have
`=lim_(x->0)(sin(4x)cos(5x))/sin(5x)`
`=lim_(x->0)sin(4x)/sin(5x)`

Now, arrange to
`=lim_(x->0)4/5*sin(4x)/(4x)*(5x)/sin(5x)`

If the factors are canceled out, this will be equal to the previous expression. The reason for this arrangement is to exploit the identity `lim_(x->0)sin(x)/x=1`.

Then, we have
`=lim_(x->0)4/5*lim_(x->0)sin(4x)/(4x)*lim_(x->0)(5x)/sin(5x)`

As `x->0`, `4x->0` and `5x->0`. Therefore, the above expression becomes
`=lim_(x->0)4/5*lim_(4x->0)sin(4x)/(4x)*lim_(5x->0)(5x)/sin(5x)`

Now we can use `lim_(x->0)sin(x)/x=lim_(x->0)x/sin(x)=1` to evaluate the individual limits:
`=4/5*1*1`
`=4/5`

We can verify this limit with a graph:
graph{sin(4x)/tan(5x) [-1,1,-1,1]}

As seen, as `x->0`, `sin(4x)/tan(5x)->4/5`.