# Find the exact limit: `lim_(x->0) (sin^4 5x)/(x^2*tan^2 3x)`

*print*Print*list*Cite

### 1 Answer

The exact value of `lim_(x->0) (sin^4 5x)/(x^2*tan^2 3x)` has to be determined.

`lim_(x->0) (sin^4 5x)/(x^2*tan^2 3x)`

=> `lim_(x->0) (sin^4 5x*cos^2 3x)/(x^2*sin^2 3x)`

Substituting x = 0 gives the indeterminate form `0/0` . Use l'Hopital's rule and substitute the numerator and denominator by their derivative.

=> `lim_(x->0) (4*5*sin^3 5x*cos 5x*cos^2 3x - sin^4 5x*2*3*sin 3x*cos 3x)/(x^2*6*sin 3x*cos 3x + 2x*sin^2 3x)`

It is seen that the numerator and denominator is equal to zero when x = 0 and l'Hopital's rule has to be applied again.

This continues till the derivative of the numerator and denominator has at least a single term in each that is not equal to 0 when x = 0

The fourth derivative of the numerator has the non-zero term `15000*cos^2 3x*cos^4 5x` .

The 4th derivative of the denominator has a non zero term `(216-648*x^2)*cos^2 3x`

Substituting x = 0 here gives `15000/216` = `625/9`

**The limit **`lim_(x->0) (sin^4 5x)/(x^2*tan^2 3x) = 625/9`