Question

evaluate lim. { tan10x }
x->0 x​

Answer 1

Answer:

Consider that tanx = sinx/cosx and that lim (x->0) sin(kx)/kx = 1 (in your case k = 10)

lim (x->0) sin(10x)/xcos(10x)

Multiply by 10/10

lim (x->0) 10sin(10x)/(10x)cos(10x) = lim (x->0) 10/cos(10x) = 10

Easy. You don't need De L'Hopital.

Number 2) Multiply by (1+cos(2x))/(1+cos(2x)) so that you have a square diff. at the top

lim (x->0) (1-cos²(2x))/(xsinx)(1+cos(2x))

Consider that cos²(2x) + sin²(2x) = 1, hence 1 - cos²(2x) = sin²(2x)

lim (x->0) sin²(2x)/(xsinx)(1+cos(2x))

You should know that sin(2x) = 2sinxcosx

lim (x->0) (2sinxcosx)²/(xsinx)(1+cos(2x)) = lim (x->0) 4sin²xcos²x/(xsinx)(1+cos(2x))

Now simplify and remember the above limit with k = 1

lim (x->0) 4sinxcos²x/x(1+cos(2x)) = lim (x->0) 4cos²x/(1+cos(2x)) = 4*1/(1+1) = 2

This is a little bit more tricky, but you still don't need De L'Hopital.