Verified answer

Divide numerator and denomintor by x,

( sin 7x / x ) / ( 1 + (tan 9x / x) ) =

( 7 sin 7x / 7x ) ( 1 + ( 9 tan 9x / 9x ) )

This approaches

7 (1) / ( 1 + 9 (1) ) = 7 / 10

Substitute x=0, we have 0/0

Apply L'Hopital's Rule. Keep on differentiating the numerator and denominator and keep on substituting x=0.

lim x-->0 sin 7x / (x + tan 9x)

= lim x--> 0 7 cos 7x / (1+ 9 sec^2 9x)

As x approaches 0, cos 7x approaches 1; sec^2 9x approaches 1;

The limit becomes 7 (1) /(1+9) = 7/10

Try plugging in 0 for x:

sin(7*0) / (0 + tan(9*0))

sin(0) / (0 + tan(0))

0 / (0 + 0)

0 / 0

That's an indeterminate form, so we can use L'Hopital's Rule. Take the derivative of the numerator and denominator:

7cos(7x) / (1 + 9sec²(9x))

Now plug in 0 for x:

7cos(7*0) / (1 + 9sec²(9*0))

7cos(0) / (1 + 9sec²(0))

7*1 / (1 + 9*1)

7 / (1 + 9)

7/10

So, the limit is 7/10.

Hope that helps :)

take derivatives

7cos(x)/(1+9sec^2(x))

as x->0

7*1/(1+ 9) = .7