Question

Find the Lim as x goes to 0 (2x/tanx) ?

Answer 1

`lim_(x->0)(2x)/tanx=2`

Explanation:

We have:

`lim_(x->0)(2x)/tanx`

When we plug `0` in the place of `x`, we get `0/0`

We use the l'Hopital's Rule, which states that:

`lim_(x->c)(g(x))/(h(x))=(g'(c))/(h'(c))` When `(g(c))/(h(c))` gives you indeterminate forms, such as `0/0` If this still gives you an indeterminate form, then you can repeat the process.

Therefore, `lim_(x->0)(2x)/tanx=(d/dx(2x))/(d/dx(tanx))`

Two things to remember:

Power rule: `x^n=nx^(n-1)` when `n` is a constant.

The `"psst"` rule.

Remove the p, then you get:

`"sst"`

which are the initials of:

`"sec,sec,tan"`

Add co to each word (with a little negative sign in the middle)

`"csc,-csc,cot"`

The derivative of the functions at the end is the product of the other two.

For example, `d/dx(secx)=secx*tanx`

Similarly, `d/dx(tanx)=sec^2x`

We now have:

`(2)/(sec^2x)`

Now replace the `x` with `0`.

`=>2/sec^2 0`

`=>2/1`

`=>2`

Therefore,

`lim_(x->0)(2x)/tanx=2`