1. ## limits problem

to find the value of $\displaystyle \lambda$ such that
$\displaystyle lim_{x\rightarrow0}[\frac{\lambda\mbox{x}^3+5x-sinx}{x^3}]=1$
some help here..

2. Originally Posted by ashes
to find the value of $\displaystyle \lambda$ such that
$\displaystyle lim_{x\rightarrow0}[\frac{\lambda\mbox{x}^3+5x-sinx}{x^3}]=1$
some help here..
No value exists. From l'Hopitals' Rule:

$\displaystyle \lim_{x \rightarrow 0}\left[ \frac{\lambda x^3+5x- \sin x}{x^3} \right] = \lim_{x \rightarrow 0} \left[ \frac{3 \lambda x^2 + 5 - \cos x}{3x^2} \right] = + \infty$ for all values of $\displaystyle \lambda$.

but after using l'hopital 'rule, can i shift everything to RHS and leave lambda on LHS,then apply the limit..cos i got $\displaystyle \lambda$=1

4. How would you "shift everything to the other side"? You have infinity to deal with which isn't a number you can operate with normally.

Also, try plugging $\displaystyle \lambda = 1$ in:

$\displaystyle \lim_{x \to 0} \frac{3(1)x^{2} + 5 - \cos x}{3x^{2}} = \left[\frac{4}{0}\right]$

ok..so is there a reason the answer is +$\displaystyle \infty$..can it be -$\displaystyle \infty$..?

6. Well, you know the numerator is positive so the only thing that could possibly make it negative is the denominator. However, you know that $\displaystyle 3x^{2} > 0$ for all x values.

7. Originally Posted by mr fantastic
No value exists. From l'Hopitals' Rule:

$\displaystyle \lim_{x \rightarrow 0}\left[ \frac{\lambda x^3+5x- \sin x}{x^3} \right] = \lim_{x \rightarrow 0} \left[ \frac{3 \lambda x^2 + 5 - \cos x}{3x^2} \right] = + \infty$ for all values of $\displaystyle \lambda$.
Why not this way?

As the denominator is x^3, finding f"'(x) / g"' (x) would result in denominator =6.

sub in the limits, equate eqn = 1 and solve for lambda.

8. Originally Posted by MilK
Why not this way?

As the denominator is x^3, finding f"'(x) / g"' (x) would result in denominator =6.

sub in the limits, equate eqn = 1 and solve for lambda.

Because applying L'hopital once is enough to get rid of indeterminate form. You can't apply it again, you would get a wrong result if you did.

9. Originally Posted by MilK
Why not this way?

As the denominator is x^3, finding f"'(x) / g"' (x) would result in denominator =6.

sub in the limits, equate eqn = 1 and solve for lambda.
l'Hopital's rule can only be used if the limit has the indeterminant form 0/0 (or oo/oo). After applying it the first time you get:

$\displaystyle \lim_{x \rightarrow 0} \left[ \frac{3 \lambda x^2 + 5 - \cos x}{3x^2} \right]$.

This no longer has the indeterminant form required for l'Hopital's rule to be applied again ...... The limit is '4/0' ......

10. Originally Posted by ashes
to find the value of $\displaystyle \lambda$ such that
$\displaystyle lim_{x\rightarrow0}[\frac{\lambda\mbox{x}^3+5x-sinx}{x^3}]=1$
some help here..
Another approach would be to substitute the opwer series around x = 0 for sin x:

$\displaystyle \lim_{x \rightarrow 0}\left[ \frac{\lambda x^3 + 5x - (x - x^3/6 + .....)}{x^3} \right] = \lim_{x \rightarrow 0}\left[ \frac{\lambda x^3 + 5x - x + x^3/6 - .....)}{x^3} \right]$

$\displaystyle = \lim_{x \rightarrow 0} \left[ \frac{ \left( \lambda + \frac{1}{6} \right) x^3 + 4x - .....}{x^3} \right] = \lim_{x \rightarrow 0} \left[ \frac{ \left( \lambda + \frac{1}{6} \right) x^2 + 4 - .....}{x^2} \right] = +\infty$

for all values of $\displaystyle \lambda$.

11. Originally Posted by mr fantastic
Another approach would be to substitute the opwer series around x = 0 for sin x:

$\displaystyle \lim_{x \rightarrow 0}\left[ \frac{\lambda x^3 + 5x - (x - x^3/6 + .....)}{x^3} \right] = \lim_{x \rightarrow 0}\left[ \frac{\lambda x^3 + 5x - x + x^3/6 - .....)}{x^3} \right]$

$\displaystyle = \lim_{x \rightarrow 0} \left[ \frac{ \left( \lambda + \frac{1}{6} \right) x^3 + 4x - .....}{x^3} \right] = \lim_{x \rightarrow 0} \left[ \frac{ \left( \lambda + \frac{1}{6} \right) x^2 + 4 - .....}{x^2} \right] = +\infty$

for all values of $\displaystyle \lambda$.
Just out of curiousity I know you know that 8/11 of the people you tell about power series substitution dont understand it because they aer in too low of a level...do you do it just in case?

12. Originally Posted by Mathstud28
Just out of curiousity I know you know that 8/11 of the people you tell about power series substitution dont understand it because they aer in too low of a level...do you do it just in case?
I'd be astonished if students at the level of l'Hospital had not been taught power series.

And if students have not been taught l'Hospital, I'd be totally astonished if they got these sorts of questions and were expected to use it .......

All too often using l'Hospital is just plain lazy. And often it cultivates a mindset that leads to its invalid use.

Using l'Hospital is like driving 50m to the convenience store instead of walking.

13. Hi

Without power series and l'Hospital :

$\displaystyle \lim_{x\to 0} \frac{ \lambda x^3+5x-\sin x}{x^3}=1 \Rightarrow \lim_{x\to 0} \frac{5-\frac{\sin x}{x}}{x^2}=1-\lambda$

As $\displaystyle \lim_{x\to 0}\frac{\sin x}{x}=1$ , (derivative of the sine taken at 0) the equality is wrong, whichever $\displaystyle \lambda$ you take...

14. if the eqn changes to :

$\displaystyle$
$\displaystyle \lim_{x\to 0} \frac{ \lambda x^3+5x-\sin 5x}{x^3}=1$

what value should lamda be?

15. Originally Posted by mr fantastic
I'd be astonished if students at the level of l'Hospital had not been taught power series.

And if students have not been taught l'Hospital, I'd be totally astonished if they got these sorts of questions and were expected to use it .......

All too often using l'Hospital is just plain lazy. And often it cultivates a mindset that leads to its invalid use.

Using l'Hospital is like driving 50m to the convenience store instead of walking.
Haha.......actually the kids at my school(discounting me ) knew about L'hopitals before they knew that $\displaystyle \int$ meant

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