Question

calculates the limit
l'hopital rule with
\[\lim_{x \rightarrow 0 } a sen(x) - x e ^{x}/x ^{^{2}}\]

Answer 1

need help....

Answer 2

@estudier

Answer 3

@amistre64
@akash123

Answer 4

wat's the 1st term?

Answer 5

arc sec(x)

Answer 6

can u write down the question again?

Answer 7

\[\lim_{x \rightarrow 0} a sen (x) e ^{x} / x ^{2}\]

Answer 8

not is arc sen
is
a sen

Answer 9

I am not aware of function ...a sen (x)

Answer 10

hmmm...
sorry
is
a sin (x)

Answer 11

then differentiate numerator and denominator..

Answer 12

i dont understand how to solve this question with rule of hopital

Answer 13

if u have to find the limit of type 0/0 0r infinity /infinity then

lim x--->a f(x)/g(x) = lim x--->a f'(x)/g'(x)

either f(a) n g(a) both tend to 0 or infinity

Answer 14

does it make sense?

Answer 15

lim x--->a f(x)/g(x) = lim x--->a f'(x)/g'(x)...this's called l'hospital's rule

either f(a) n g(a) both tend to 0 or infinity

Answer 16

is it fine?

Answer 17

e.g lim x--->0 asinx (e^x)/x^2 ( form of 0/0) ...right?

Answer 18

lim x--->0 asinx (e^x)/x^2
= lim x--->0 a[ e^x sinx + e^x cosx]/ 2x

Answer 19

ok

Answer 20

have u written down the question correctly?

Answer 21

rite

Answer 22

lim x--->0 asinx (e^x)/x^2 ...is this the right question?

Answer 23

if yes then
lim x--->0 asinx (e^x)/x^2
= lim x--->0 a[ e^x sinx + e^x cosx]/ 2x
= lim x--->0 a/ 2x
and limit does not exist

Answer 24

yes

Answer 25

do u know the answer?

Answer 26

no :(

Answer 27

\[\Large \lim_{x \rightarrow 0} \frac{asin(x)e^{x}}{x^2}\] is this is your question ?

Answer 28

yesss

Answer 29

limit doesn't exist for this question

Answer 30

might be ...it should be sin^2x instead of sinx

Answer 31

but now u know the l'hospital rule...u can try other questions

Answer 32

ok by applying the limits you get 0/0 form which means this question is condiate for L Hopital rule

in this rule you differentiate both numerator and denominator
by taking derivative of numerator we have by product rule of differentiation
\[\Large \frac{d}{dx}(asin(x)*e^x)=a \frac{d}{dx}(\sin(x)*e^x)=a[\cos(x)e^x+\sin(x)e^x)]\]
the derivative of denominator is 2x so
\[\Large \lim_{x \rightarrow 0} \frac{a[\cos(x)e^x+\sin(x)e^x]}{2x}\]
now apply the limit
\[\Large \frac{a[\cos(0)e^0+\sin(0)e^(0)]}{2(0)}\]
where
\[\Large \cos(0)=1\]
\[\Large \sin(0)=0\]
\[\Large e^{0}=1\]
let me know what u get after the simplification of above ..

Answer 33

1/0

Answer 34

you missed a
it is a/0 so it means \[\Large \frac{a}{0}=\infty\] should be the answer .

Answer 35

yes...thanks :)))

Answer 36

welcome :)

Answer 37

:))

Answer 38

Sir:)