Question

limits question:
I will solve it just check my answer :)

Answer 1

\[\lim_{x \rightarrow 0}\ e^{x^2}-cosx/\sin^2x\ = 3/2\]

Answer 2

is the answer 3/2 correct ?

Answer 3

@Michele_Laino please check :)

Answer 4

is your function like below:
\[\frac{{{e^{{x^2}}} - \cos x}}{{{{\left( {\sin x} \right)}^2}}}\]

Answer 5

yes:)

Answer 6

Yep, that's what I got.

Answer 7

I can rewrite your function as below:
\[\frac{{{e^{{x^2}}} - \cos x}}{{{{\left( {\sin x} \right)}^2}}} = \frac{{{e^{{x^2}}} - \cos x}}{{{x^2}}}\frac{1}{{\frac{{{{\left( {\sin x} \right)}^2}}}{{{x^2}}}}} = \frac{{{e^{{x^2}}} - \cos x}}{{{x^2}}}\frac{1}{{{{\left( {\frac{{\sin x}}{x}} \right)}^2}}}\]

Answer 8

as we know, we have:
\[\mathop {\lim }\limits_{x \to 0} \frac{1}{{{{\left( {\frac{{\sin x}}{x}} \right)}^2}}} = 1\]

Answer 9

whereas the second limit, using the L'Hopital rule, is:
\[\large \mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - \cos x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{2\left( {{e^{{x^2}}} + 2{x^2}{e^{{x^2}}}} \right) + \cos x}}{2} = \frac{3}{2}\]
so their product is 3/2
So your answer is right!

Answer 10

that is correct i aggree

Answer 11

thank you so much everyone
thank you @Michele_Laino for taking time to explain :)

Answer 12

well i used the other way to solve it :)

Answer 13

please, note that for second limit, we have to apply the L'Hopital rule two times
Thank you :) @rvc

Answer 14

here is what i did
\[e^{x^2}-1+1-\cos/x^2\]

Answer 15

then taking them seperately
for e term i get 1 and the cos term i get 0.5
adding them i get 3/2 :)
i dont know how to use that rule
may u please explain @Michele_Laino

Answer 16

ok! it looks a good way!

Answer 17

thank you :)

Answer 18

In order to apply the L'Hopital rule, we have this undetermined form:
\[\large \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \frac{0}{0}\]

Answer 19

under that condition, we can state the requested limit exists, and its value is:
\[\large l = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f'\left( x \right)}}{{g'\left( x \right)}}\]

Answer 20

if, we get:
\[\large \mathop {\lim }\limits_{x \to {x_0}} \frac{{f'\left( x \right)}}{{g'\left( x \right)}} = \frac{0}{0}\]
then we can apply the L'Hopital rule another time to a ratio:
\[\large \frac{{f'\left( x \right)}}{{g'\left( x \right)}}\]
and so on

Answer 21

so can u try this problem with this rule just as an example if it doesnt take your time :)

Answer 22

ok! We have the subsequent steps:

Answer 23

thank you
lets start :)

Answer 24

\[\large \begin{gathered}
\mathop {\lim }\limits_{x \to 0} \frac{{{e^{{x^2}}} - \cos x}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \frac{{2x{e^{{x^2}}} + \sin x}}{{2x}} = \hfill \\
\hfill \\
\mathop {\lim }\limits_{x \to 0} \frac{{2\left( {{e^{{x^2}}} + 2{x^2}{e^{{x^2}}}} \right) + \cos x}}{2} = \frac{3}{2} \hfill \\
\end{gathered} \]
since the first derivative and the second derivative of the numerator, are respectively:
\[\Large {2x{e^{{x^2}}} + \sin x}\]
\[\Large {2\left( {{e^{{x^2}}} + 2{x^2}{e^{{x^2}}}} \right) + \cos x}\]
whereas, the first derivative and the second derivative of the denominator, are, respectively:
\[\Large {2x}\]
\[\Large 2\]

Answer 25

okay

Answer 26

:)

Answer 27

hey wait how 3/2?

Answer 28

okay!
thank you!

Answer 29

thank you!

Answer 30

*