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L'Hospital's Rule

renyikouniao

Member
Jun 1, 2013
41
Question:Determine 1.Lim(x->infinity)e^(-x)coshx
2.Lim(x->1)[(e^x-1)/In(x)]sinhx

For 1.=Lim(x->infinity)e^(-x)/(1/coshx) so the top and bottem both go to 0.then apply the rule. Repeat this procedure 3 times the top and bottem still both go to 0,so I am stuck.

2.Circumstance same as 1.I apply the rule 2 times and both the top and the bottem go to infinity.
 
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
1.) I would write the expression as:

\(\displaystyle \frac{e^x+e^{-x}}{2e^x}=\frac{1+e^{-2x}}{2}\)

Now you have a determinate form.

2.) Substituting 1 for $x$ gives you a constant divided by zero...so check the one-sided limits.
 

renyikouniao

Member
Jun 1, 2013
41
1.) I would write the expression as:

\(\displaystyle \frac{e^x+e^{-x}}{2e^x}=\frac{1+e^{-2x}}{2}\)

Now you have a determinate form.

2.) Substituting 1 for $x$ gives you a constant divided by zero...so check the one-sided limits.
Thank you for the reponse.
For part 2. (e^x-1)/Inx is one part,sinhx is another part.They both go to infinity as x approches 1.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle \sinh(1)=\frac{e-\frac{1}{e}}{2}=\frac{e^2-1}{2e}\approx1.1752011936438014\)
 

renyikouniao

Member
Jun 1, 2013
41
\(\displaystyle \sinh(1)=\frac{e-\frac{1}{e}}{2}=\frac{e^2-1}{2e}\approx1.1752011936438014\)
Thank you very much!:D How about I think it this way: (e^x-1/In(x))/(1/sinhx)
(e^x-1/In(x)) this whole thing approches negative infinity and 1/sinhx approches negative infinity as x approches 1
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
But \(\displaystyle \lim_{x\to1}\frac{1}{\sinh(x)}\ne-\infty\)...instead we have:

\(\displaystyle \lim_{x\to1}\frac{1}{\sinh(x)}=\frac{2e}{e^2-1}\)

So we know:

\(\displaystyle \lim_{x\to1}\left(e^x-1 \right)\sinh(x)=\frac{(e-1)\left(e^2-1 \right)}{2e}\)

which is a constant, let's call it $k$, which means the original limit may be written

\(\displaystyle k\lim_{x\to1}\frac{1}{\ln(x)}\)

We know the natural log function in the denominator is approaching zero, but do we have:

\(\displaystyle \lim_{x\to1^{-}}\frac{1}{\ln(x)}=\lim_{x\to1^{+}}\frac{1}{\ln(x)}\) ?