Indeterminate forms and L'Hospital's Rule

[asimon2005]

48. Lim (cscx- cotx)
x-0

52. lim (xe^1/x -x)
x-?

 

[galactus]

48. \lim_{x\to 0} (csc(x)- cot(x))[/tex]

Note that \(\displaystyle csc(x)-cot(x)=tan(\frac{x}{2})\)

\(\displaystyle 52. \lim_{x\to \infty}(xe^{\frac{1}{x}}-x)\)

For this one, we could make a sub by letting t=1/x

This changes the limit to t-->0 and we have

\(\displaystyle \lim_{t\to 0}\frac{e^{t}-1}{t}\)

Now, by L'Hopital, we get \(\displaystyle \lim_{t\to 0}e^{t}\)

Now, what is the limit as t-->0?.

We can do this various ways without L'Hopital.

Express \(\displaystyle e^{t}\) as a derivative using the defintion of a derivative:

\(\displaystyle \frac{d}{dt}[e^{t}]=\lim_{h\to 0}\frac{e^{0+h}-e^{0}}{h}=\lim_{h\to 0}\frac{e^{h}-1}{h}=\)

Check this out: http://www.ies.co.jp/math/java/calc/lim_e/lim_e.html


Here 's a cool way using algebra and no L'Hopital.

Let \(\displaystyle \lim_{t\to 0}\frac{e^{t}-1}{t}\)

Let \(\displaystyle u=\frac{1}{e^{t}-1}\)

\(\displaystyle \frac{1}{u}=\frac{e^{t}-1}{1}\)

\(\displaystyle 1+\frac{1}{u}=e^{t}\)

\(\displaystyle t=ln(1+\frac{1}{u})\)

\(\displaystyle \lim_{u\to \infty}\frac{1}{uln(1+\frac{1}{u})}\)

\(\displaystyle \lim_{u\to \infty}\frac{1}{ln((1+\frac{1}{u})^{u})}\)

But, note what is inside the parentheses is the limit for e.

We have: \(\displaystyle \frac{1}{ln(e)}=\)

3 different ways to do the same limit. L'Hopital is easiest though.

 

[Deleted member 4993]

48. Lim (cscx- cotx)
x-0

If you have to use L'Hospital then

\(\displaystyle csc(x) - cot(x) = [1- cos(x)]/sin(x)\)

52. lim (xe^1/x -x)
x-?