Find the lim of (sin(x^3+x^2-x)+sinx)/x as x approaches 0

[iremos]

Hi,

I need some help in solving this question,

Find lim(x→0) ⁡(sin⁡(x³+x²-x)+sin⁡x)/x without using derivative given that lim(u→0) sinu/u = 1

here is what i have work out so far,

= sinx(x²+x-1)/x +sinx/x
= sinx(x²+x-1)/x + 1

I would like to ask, if i apply L'Hopital's rule, does it means I'm using derivative?

 

[Deleted member 4993]

Hi,

I need some help in solving this question,

Find lim(x→0) ⁡(sin⁡(x³+x²-x)+sin⁡x)/x without using derivative given that lim(u→0) sinu/u = 1

here is what i have work out so far,

= sinx(x²+x-1)/x +sinx/x
= sinx(x²+x-1)/x + 1

I would like to ask, if i apply L'Hopital's rule, does it means I'm using derivative? ........ yes you are using derivatives while using L'Hopital's rule

[sin⁡(x³+x²-x)+sin⁡x)]/x

= sin⁡(x³+x²-x)/x + sin⁡(x)/x

=sin⁡(x³+x²-x)/x +sin⁡(x)/x

=sin⁡(x³+x²-x)/[x(x²+x -1)] * (x²+x -1) + sin⁡(x)/x

Now continue.....

 

[Steven G]

Hi,

sinx(x²+x-1)/x +sinx/x
= sinx(x²+x-1)/x + 1

No No No! sinx/x does not equal 1. No!!!!

Possibly the limit as x->0 of sinx/x =1 but sinx/x does not equal 1.

 

[Otis]

Possibly the limit as x->0 of sinx/x =1

Hmm, the exercise is a limit, with x approaching zero.

And the limit of the form sin(x)/x is given in the OP.

given that lim(u→0) sinu/u = 1

 

[iremos]

[sin⁡(x³+x²-x)+sin⁡x)]/x

= sin⁡(x³+x²-x)/x + sin⁡(x)/x

=sin⁡(x³+x²-x)/x +sin⁡(x)/x

=sin⁡(x³+x²-x)/[x(x²+x -1)] * (x²+x -1) + sin⁡(x)/x

Now continue.....

Thank you for your help! Sorry for the late reply. I was busy with work.

I tried to work it out,

= sin⁡(x³+x²-x)/[x(x²+x -1)] * (x²+x -1) + sin⁡(x)/x

= sinx(x²+x-1)/[x(x²+x -1)] * (x²+x -1) + sin⁡(x)/x

= sinx/x * (x²+x -1) + sin⁡(x)/x

= 1 *(x²+x -1) + 1

= 0²+0-1+1

= 0

I'm not sure if cancelling out (x²+x-1) makes any sense, please let me know if it's incorrect.

 

[SAQLAIN AHMAD]

Hi,

I need some help in solving this question,

Find lim(x→0) ⁡(sin⁡(x³+x²-x)+sin⁡x)/x without using derivative given that lim(u→0) sinu/u = 1

here is what i have work out so far,

= sinx(x²+x-1)/x +sinx/x
= sinx(x²+x-1)/x + 1

I would like to ask, if i apply L'Hopital's rule, does it means I'm using derivative?

Applying SinC+SinD formula,we get
L=lim(x→0)((2sin((x^3+x^2)/2)cos((x^3+x^2-2x)/2))/x
=lim(x→0)((2sin((x^3+x^2)/2)cos((x^3+x^2-2x)/2))*0.5(x^3+x^2)/(0.5(x^3+x^2)*x)
Here lim(x→0)(sin((x^3+x^2)/2)/0.5(x^3+x^2)=1.
So L=lim(x→0)2cos((x^3+x^2-2x)/2)*(0.5(x^3+x^2)/x=lim(x→0)2cos((x^3+x^2-2x)/2)*(0.5(x^2+x)=1*0=0

OR
Simply apply L-Hospital rule,
L=lim(x→0)cos(x^3+x^2-x)(3x^2+2x-1)+cosx=-1+1=0

 

[SAQLAIN AHMAD]

Thank you for your help! Sorry for the late reply. I was busy with work.

I tried to work it out,

= sin⁡(x³+x²-x)/[x(x²+x -1)] * (x²+x -1) + sin⁡(x)/x

= sinx(x²+x-1)/[x(x²+x -1)] * (x²+x -1) + sin⁡(x)/x

= sinx/x * (x²+x -1) + sin⁡(x)/x

= 1 *(x²+x -1) + 1

= 0²+0-1+1
= 0

I'm not sure if cancelling out (x²+x-1) makes any sense, please let me know if it's incorrect.

The Method you have applied is incorrect,although the answer turned out to be correct.You just cannot cancel (x^2+x-1) like that

 

[Deleted member 4993]

Thank you for your help! Sorry for the late reply. I was busy with work.

I tried to work it out,

= [sin⁡(x³+x²-x)/[x(x²+x -1)] * (x²+x -1) + sin⁡(x)/x

= [sin⁡(x³+x²-x)/(x³+x²-x)] * (x²+x -1) + sin⁡(x)/x

Now consider limit (x→0) and limit [(x³+x²-x)→0]

= 1 * (0²+0 -1) + 1

= 0

See the corrections made above in red.

 

[Deleted member 4993]

Hi,

I need some help in solving this question,

Find lim(x→0) ⁡(sin⁡(x³+x²-x)+sin⁡x)/x without using derivative given that lim(u→0) sinu/u = 1

I would like to ask, if i apply L'Hopital's rule, does it means I'm using derivative?

While using L'Hopital's rule - you have to use derivative.

So for this (restricted) problem you CANNOT use L'Hopital's rule.

However, you can check your answer using L'Hopital's rule.

 

[pka]

Find lim(x→0) ⁡(sin⁡(x³+x²-x)+sin⁡x)/x without using derivative given that lim(u→0) sinu/u = 1

This is one of the easiest questions I seen posted here. It is just simple algebra.
\(\displaystyle \displaystyle{{\lim _{x \to 0}}\frac{{\sin \left( {{x^3} + {x^2} - x} \right)}}{x} = {\lim _{x \to 0}}\frac{{\sin \left( {{x^3} + {x^2} - x} \right)}}{{{x^3} + {x^2} - x}}\frac{{{x^3} + {x^2} - x}}{x}\\ = {\lim _{x \to 0}}\frac{{\sin \left( {{x^3} + {x^2} - x} \right)}}{{{x^3} + {x^2} - x}}\left( {{x^2} + x - 1} \right) \to \left( 1 \right)\left( { - 1} \right) = - 1}\)

 

[SAQLAIN AHMAD]

See the corrections made above in red.

Ya now this is perfectly fine

 

[iremos]

I'm sorry that my question was a bit silly. I have not been studying for almost 8 years and I'm trying to get my degree now, please bear with me . this question is one of the past admission exam questions.

Thank you all for your help! appreciate it a lot.