Question

Lim (x approaches to 0) of (e^2x -e^x)/sinx Don't understand how to solve this. Don't know how to get 3 (the answer) from this, please help.

Answer 1

apply l hospitals rule...as it is in 0/0 form

Answer 2

differentiat the numerator first and then denominator until 0/0 form vanishes and gets reduced to some number.

Answer 3

have u undrstood?if njot,tell me.i will explain it to u step by step.

Answer 4

Can you please explain it?

Answer 5

ok first of all tell me do u know differentiation?

Answer 6

pls rply fast

Answer 7

I tried dividing the denominator by x in order to get rid of sin and I got an x for the denominator. But I don't know what to do from there on.

Answer 8

no no u dont need to.here u have to apply l hodpitals rule.....ok first differentiate the numerator and tell me wt u got.

Answer 9

Do you mean this? Lim(x->0) ( e^2x )/sinx - (e^-x) /sinx

Answer 10

in numerator we have e^2x-e^x
by differentiating,we get
2e^2x-e^x

Answer 11

have u undrstood dis diffrentiation?

Answer 12

Not really

Answer 13

in denominator we have sinx.now we have to differentiate it as well,
so we get cosx.
now,
[2e^2x-e^x]/cosx

Answer 14

now put the value of limit,i.e.x=o in it u will get the ans.

Answer 15

differentiation of e^ax=a*e^ax do u know dis?

Answer 16

Yes

Answer 17

If you don't know the L'Hospitals rule try this,
\[\lim \limits_{x\to0} \frac{e^{2x}-e^x}{\sin x}=\lim \limits_{x\to0} \frac{e^x(e^{x}-1)}{\sin x}\\=\lim \limits_{x\to0} e^x\frac{(e^{x}-1)}{x}\frac{1}{\left(\frac{\sin x}{x}\right)}\\=\left(\lim \limits_{x\to0}e^x\right)\left( \lim \limits_{x\to0}\frac{(e^{x}-1)}{x}\right)\frac{1}{\left(\lim \limits_{x\to0}\frac{\sin x}{x}\right)}\]

Answer 18

Which is 1* lim(x->0) e^x -1)/x But where do I go from this?Thr denominator can't be zero

Answer 19

Remember:
\[e^x=1+\sum\limits_{i=1}^\infty \frac{x^n}{n!}\\
\frac{e^x-1}{x}=\sum\limits_{n=1}^\infty \frac{x^{n-1}}{n!}=1+\sum\limits_{n=2}^\infty\frac{x^{n-1}}{n!}\\
\lim\limits_{x\to0}\frac{e^x-1}{x}=1+\lim\limits_{x\to0} \sum\limits_{n=2}^\infty\frac{x^{n-1}}{n!}=1+0=1\]

Answer 20

Ok thanks :)

Answer 21

l hospital is easier.if u learn abt l hospital u will know it dat its a far better process.

Answer 22

@Mathstudent97
madrockz is right ...when you solve the complex question then l hospital rule is very easy and time saving compare to another way..if you dont know differentiation then learn it and after it learn l hospital rule then you will observe that l hospital rule is very easy for 0/0 form ..and @madrockz thanks....

Answer 23

Thanks I will :)