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