Question

Find lim_{x rightarrow 0} (tan 3x^{2} + sin ^{2} 5x) / x ^{2}
i'd tried applying L'Hospital Rule twice, and i'd obtained -25 as the answer. is there any other method to solve this question? or if there aint any, is my answer correct?

Answer 1

\[\frac{\tan(3x^2)+\sin^2(5x)}{x^2}\]?

Answer 2

\[\lim_{x \rightarrow 0} (\tan 3x^{2} + \sin^{2} 5x) / x ^{2} \]

Answer 3

yes

Answer 4

8, I believe

Answer 5

I get a mess...

Answer 6

\[\lim x \rightarrow0 {tanx \over x} = 1 and \lim x \rightarrow 0{sinx \over x} = 1 \]
also when x tends to zero x^2 tends to zero

Answer 7

so you can use that limit to get the answer as\[\lim x \rightarrow 0 {\tan (3x^2) \over x^2 } = \lim x \rightarrow0(3{\tan(3x^2) \over 3x^2}) = 3\]

Answer 8

Break it up into parts. First, the limit of
\[ \frac{\tan(3x^2)}{x^2} \]
As the limit goes to zero, tangent behaves just like sine. The Maclaurin series would therefore begin with 3x^2. The x^2 terms cancel, leaving a 3.
Then,
\[\frac{\sin^2(5x)}{x^2} = \left(\frac{\sin(5x)}{x}\right)^2 \]
the limit of which is 25. So I should revise my answer, I believe it should be 28.

Answer 9

o_O?

Answer 10

wolf agrees with you Jemurray3!

Answer 11

umm so where did i go wrong?

Answer 12

Hooray! @Akshay I don't think you did. The limit of the tangent part is 3, as you say, and the limit of the sine part is 5^2 = 25

Answer 13

oh yea! i am still to solve the sine part -_-

Answer 14

my bad :P

Answer 15

@.@ wow means i need to work harder on calculus == anyway i handed in the answer last week. not expecting to obtain any solutions from my lecturer, that's why i asked here. he's just touching the tip of the topic... so conclusion is working out separately to find the limits?

Answer 16

Also, I should apologize.... that wasn't a very academic answer. Filled with my own little shortcuts and intuitive reasoning. To be more precise, notice that
\[ \frac{\tan(3x^2)}{x^2} = 3\frac{\tan(3x^2)}{3x^2} = 3\frac{\tan(u)}{u} \]
The limit of that as u -> 0 is 3.

Answer 17

Just like Akshay said.
Then,
\[ \frac{sin^2(5x)}{x^2} = \left( 5\frac{\sin(5x)}{5x}\right)^2 = 25\left(\frac{\sin(5x)}{5x}\right)^2\]
\[ = 25\left(\frac{\sin(u)}{u} \right)^2 \]
The limit of which as u -> 0 is 25.

Answer 18

owh. thank you all very much. need some time to read up. lolx.

Answer 19

off topic: im 20yo this year, am i too stupid for not knowing how to solve this question? ==

Answer 20

Age is just a number, it cant define what you should know and what you shouldnt

Answer 21

Also stupidity is a relative concept, depends on who looks at it in what manner. Even if a 50 year old didn't know this , i would not tell he is stupid. The fact that you know how to do it now is of more importance

Answer 22

mmm could be my foundation on trigo and calculus aint strong enough. i always hated trigo lol

Answer 23

lol the foundation can be built anytime :P