Question

The value of \[\lim_{x \rightarrow 0} \cfrac{\int_{0} ^{x^2} \sec^2 t dt }{x \sin x} = ? \]

Answer 1

@vishweshshrimali5

Answer 2

First solve the integration part,

the integration of sec^2 x dx is ?

Answer 3

tanx + C (it is definite then remove C)

Answer 4

Good

Answer 5

Or you can use LH rule also

Answer 6

Hopital will work also

Answer 7

so if I use LH rule... then the numerator will become \(sec^2 t dt\) ?

Answer 8

No

Answer 9

Maclaurin

Answer 10

Remember this:

*sigh* I don't know latex codes...

Answer 11

Okay.. I thought integration is anti-derivative and so, if I differentiate the integral, it will give the function inside it.

Answer 12

chain rule - bounds matter !

Answer 13

\[\large \dfrac{d}{dx} \int \limits_{0}^{x^2} f(t) dt = f(x^2)\times (x^2)' \]

Answer 14

\[\cfrac{d}{dx}\int\limits_{f(x)}^{g(x)}\alpha(t).dt = \alpha(g(x)).g'(x) - \alpha(f(x)).f'(x)\]

Answer 15

Huhh I ultimately wrote it...

Answer 16

Well @ganeshie8 has put the values in the general one. You can check it @mathslover

Answer 17

Yeah.. so I can write what @ganeshie8 wrote as :

\(f(x^2) \times 2x\) And for the question :

\(\sec^2 (x^2) \times 2x\) ... will become the numerator? is that right?

Answer 18

Yeah

Answer 19

You could have obtained the same thing, if you had first solved the integral and then applied the LH rule.

Answer 20

SO, I get :
\[ \lim_{x\rightarrow 0} \cfrac{\sec^2 (x^2) \times 2x}{ \cos x}\]

Answer 21

Right?

Answer 22

Oh no... sorry

Answer 23

\(\sin x + x \cos x\) will be the denominator, right?

Answer 24

yeah

Answer 25

apply LH rule again

Answer 26

Okay.. give me one minute :-)

Answer 27

take your time.

Answer 28

Its a little bit complicated :/

Answer 29

\[\lim_{x \rightarrow 0} \cfrac{\int_{0} ^{x^2} \sec^2 t dt }{x \sin x} = \lim_{x \rightarrow 0} \cfrac{\tan(x^2) }{x \sin x} = \lim_{x \rightarrow 0} \cfrac{\dfrac{\tan (x^2)}{x^2} }{\dfrac{\sin x}{x}} \]

Answer 30

the first idea was simple ^ :)

Answer 31

@ganeshie8 yes I got 1 from there already

Answer 32

yeah that's a better method.

Answer 33

Also, if you had obtained sin x + xcos x in the denominator then divide both numerator and denominator by x and apply this :

lim x--> 0 sinx/x = 1

Answer 34

2 sec^2 (t) tan (t)

Answer 35

@vishweshshrimali5

Answer 36

Yeah.. I got how you didn't use LH rule again. But, any way to go on from there? ^

Answer 37

you differentiated the numerator wrongly

Answer 38

o.O Yeah..

Answer 39

remembe derivative sec (x) is sec(x)tan(x)

Answer 40

I thought that there is sin x in the denominator :(

Answer 41

Bhaiya, is the numerator right?

Answer 42

Nope

Answer 43

Neither numerator nor denominator are right

Answer 44

Fine got it..

Answer 45

first note the result of first application of LH rule

Answer 46

this is what I have..

Answer 47

I would advise you not to use this method.

\(\color{blue}{\text{Originally Posted by}}\) @vishweshshrimali5
Also, if you had obtained sin x + xcos x in the denominator then divide both numerator and denominator by x and apply this :

lim x--> 0 sinx/x = 1
\(\color{blue}{\text{End of Quote}}\)

See the limit you obtained after applying the LH Rule for first time. Divide the numerator and denominator by x. And solve. You will get your answer

Answer 48

Yep.. I understand that. But, I was just checking my differentiation skills

Answer 49

the differentiation done by me there is right or wrong? can u please tell this?