## MarcLeclair 2 years ago 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

1. satellite73

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

2. satellite73

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

3. satellite73

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

4. satellite73

oh i see still a problem

5. satellite73

maybe flip the other way

6. sirm3d

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

7. MarcLeclair

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

8. MarcLeclair

or am I just not understanding something ?

9. satellite73

no @sirm3d as it! first limit is 1

10. satellite73

*has it

11. MarcLeclair

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

12. MarcLeclair

Thanks to both of you