Question

find the limit

Answer 1

\[\lim_{x \rightarrow 0}\frac{ \tan(x) -x}{ x^2 \tan(x) }\]

Answer 2

I expect you have to apply L'Hopital's rule at least two times, from what I did quick in my head. Sorry for not giving you more instruction. You're familiar with L'Hopital's Rule, right?

Answer 3

Yeah! lemme try

Answer 4

I think one time Differentiating is enough.

Answer 5

I don't think so. I think if you apply the limit again, you'll get 0/0 again.

Answer 6

\[\lim_{x \rightarrow 0} \frac{\sec^2(x) - 1}{x^2 \sec^2(x) + 2xtan(x)}\]

Answer 7

I'm thinking for another method to do this.

Answer 8

;-; Probably would be helpful. I'm not seeing a good way.

Answer 9

Hmm, well, LH rule is the best for such questions (0/0 form .. ) ....! But, I always love alternative methods ;)

Answer 10

cant do it.
Can some one help. :(

Answer 11

Saumya, where are you stuck?

Answer 12

Are you okay with first time LH rule? :

\(\color{blue}{\text{Originally Posted by}}\) @iPwnBunnies
\[\lim_{x \rightarrow 0} \frac{\sec^2(x) - 1}{x^2 \sec^2(x) + 2xtan(x)}\]
\(\color{blue}{\text{End of Quote}}\)

Answer 13

Yea

Answer 14

divide numerator and denominator by x

Answer 15

@hartnn yeah, I tried that.

Answer 16

I know, limit (x tends to zero) tanx/x = 1

Answer 17

Why don't I know that. ;-;

Answer 18

you are still left with 0/0 ...

limit(tanx/x - 1)/(xtanx)

Answer 19

aah!

Answer 20

\(\lim {x\rightarrow 0} \cfrac{(\tan x )/ x - 1}{x \tan x}\)
again 0/0 ...

Answer 21

LH rule will be the only solution, I think.

@hartnn - any thoughts?

Answer 22

</3

Answer 23

let us forget tan (x) for an instant.

we have limit ( x tends to zero) ( y- x) / (x^2 y)

Answer 24

yes. then?

Answer 25

No saumyaa.... not now! We're just trying to find another method (possibly easier one..)

Till then, tell me what is the differentiation of sec^2 (x) w.r.t x

Answer 26

2 secx * sec x tan x

Answer 27

Hmm! Right. SImilarly find the derivatives of the functions in the denominator.

Answer 28

Hel anyone maybe?

Answer 29

You'd have to find the derivatives of the bottom. ;-;
Derivative of x^2*sec^2(x)

Answer 30

but still. i am not able to do it.
Answer is 1/3 Btw.

Answer 31

\[\lim_{x \rightarrow 0} \frac{2\sec^2 x \tan x}{x^2*2 \sec^2(x) \tan(x) + 2xsec^2(x) + 2xsec^2(x) + 2\tan(x)}\]

Answer 32

Yeah! I got this but do we have to put the limit or simplify further?

Answer 33

Oh my. ._. This limit is ugly.

Answer 34

tanX common for Numerator and denominator?

Answer 35

It's not in all terms in the denominator. Give me a minute.

Answer 36

Heh, from the solution I just saw. L'Hopital's was applied 3 times.