Question

Find the limits:
lim tan 3t/4t
t→0

lim tan 2x / sin 5x
x→0

Answer 1

Do u know L'Hopitals rule?

Answer 2

ok so what it says is:\[\lim_{x \rightarrow a}\frac{ f(x) }{ g(x) }\]=\[\frac{ \lim_{x \rightarrow a}f'(x) }{ \lim_{x \rightarrow a}g'(x) }\]

Answer 3

only if the denominator of the limit when you plug a in is undefined

Answer 4

So basically you take the derivative of each function and then take the limits and divide them

Answer 5

so for your first problem:\[\lim_{t \rightarrow 0}\frac{ \tan(3t) }{ 4t }\]=\[\frac{ \lim_{t \rightarrow 0}3\sec^2(3x) }{ \lim_{t \rightarrow 0}4 }\]

Answer 6

th ederivative of tan(3x) is 3sec^2(3x) using the chain rule

Answer 7

the derivative of 4t is 4

Answer 8

that means the value of your first limit is 3/4

Answer 9

Similarly your second limit is:\[\lim_{x \rightarrow 0}\frac{ \tan(2x) }{ \sin(5x) }\]=\[\frac{ \lim_{x \rightarrow 0}2\sec^2(2x)}{ \lim_{x \rightarrow 0}5\cos(5x) }\]

Answer 10

can you try to do it by yourself for the second limit?

Answer 11

So basically what I did is took the derivatives of the numerator and denominator and then took the limits of the derivatives and put them over each other.