Question

i think i am still confused with this A and B things what about If A and B are positive constants, then find lim x-> 0 tan(Ax^2)/(Bx)^2?

Answer 1

i got answer A/ B^2
any comment?

Answer 2

Hello!

Answer 3

\(\color{#000000}{ \displaystyle \lim_{x\to0} \frac{\tan(Ax^2)}{Bx} }\)

By applying L'Hospital's Rule I get:

\(\color{#000000}{ \displaystyle \lim_{x\to0} \frac{2Ax\sec^2(Ax^2)}{B} }\)

You can re-write the limit, but answer is pretty obvious,

\(\color{#000000}{ \displaystyle \left(\lim_{x\to0} \frac{2A\sec^2(Ax^2)}{B} \right)\times \lim_{x\to0} x }\)

So, yes you still end up with ....

Answer 4

have you learned L'Hospital's rule?

Answer 5

In fact, for all \(n>1\).

\(\color{#000000}{ \displaystyle \lim_{x\to0} \frac{\tan(Ax^n)}{Bx} =0 }\)

I can prove this by applying L'Hospital's rule:

\(\color{#000000}{ \displaystyle \lim_{x\to0} \frac{nAx^{n-1}\sec^2(Ax^n)}{B} }\)

And then I get,

\(\color{#000000}{ \displaystyle \left( \lim_{x\to0} \frac{n\sec^2(Ax^n)}{B} \right)\times \lim_{x\to0} Ax^{n-1} =\frac{n}{B}\times 0=0 }\)

Answer 6

If n<1, then the limit would diverge.
If n=1, then following the same derivation the limit is n/B.

Answer 7

> i got answer A/ B^2

I agree. @kws942002