MHB Answer Lim(x->infinity)e^(-x)coshx and Lim(x->1)[(e^x-1)/In(x)]sinhx

[renyikouniao]

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.

 

[MarkFL]

1.) I would write the expression as:

$$\frac{e^x+e^{-x}}{2e^x}=\frac{1+e^{-2x}}{2}$$

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]

1.) I would write the expression as:

$$\frac{e^x+e^{-x}}{2e^x}=\frac{1+e^{-2x}}{2}$$

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]

$$\sinh(1)=\frac{e-\frac{1}{e}}{2}=\frac{e^2-1}{2e}\approx1.1752011936438014$$

 

[renyikouniao]

$$\sinh(1)=\frac{e-\frac{1}{e}}{2}=\frac{e^2-1}{2e}\approx1.1752011936438014$$

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]

But $$\lim_{x\to1}\frac{1}{\sinh(x)}\ne-\infty$$...instead we have:

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

So we know:

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

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

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

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

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