MarcLeclair Group Title Use L'hospital rule to find the following limit: Lim x-->0+ 6(tan(4x))^x I did this: Lim x-->0+ xln 6(tan(4x)) which is the same as: ln 6(tan(4x)) / (1/x) Take the derivative: (1/6tan(4x))) * 6(sec^2(4x))(4) / (-1/x^2). But I can't find a way to cancel out my denominator :/ the answer isn't infinity either one year ago one year ago

1. satellite73 Group Title

leave the six out of it, put it in at the end

2. satellite73 Group Title

$x\ln(\tan(4x)$ $\frac{\ln(\tan(4x))}{\frac{1}{x}}$etc

3. satellite73 Group Title

$-4x^2\csc(4x)\sec(4x)$ i think is the next step

4. satellite73 Group Title

oh i see still a problem

5. satellite73 Group Title

maybe flip the other way

6. sirm3d Group Title

$\large= \frac{ 4x }{ \sin 4x }\frac{ -x }{ \cos 4x }$ there is no problem

7. MarcLeclair Group Title

your sin will make the denominator 0 even if multiplied :/

8. MarcLeclair Group Title

or am I just not understanding something ?

9. satellite73 Group Title

no @sirm3d as it! first limit is 1

10. satellite73 Group Title

*has it

11. MarcLeclair Group Title

0ooh take the limit of each... stupid me.

12. MarcLeclair Group Title

Thanks to both of you