# Thread: l'hopitals rule checks and help

1. ## l'hopitals rule checks and help

1: lim(x-->0) sin(sinx)/sinx...i have solved using l'hopitals rule to get 1

2: lim(x-->infinity) ln(x)/(x^2) i have got a value of 0

3: The size of an animal’s pupils expand and contract depending on the amount of light available. Let f(x) = (160x^(-0.4)+90)/(8x^(-0.4)+10)
be the size in mm of the pupils at light intensity x. Find lim(x-->0+) and lim(x-->inf) for f(x), and argue that these represent the largest and smallest possible sizes of the pupils

Can someone please check my answers for 1 and 2 and give me some help for question 3?

P.S sorry about the bad notation for the equation and limits

2. Originally Posted by deragon999
1: lim(x-->0) sin(sinx)/sinx...i have solved using l'hopitals rule to get 1

2: lim(x-->infinity) ln(x)/(x^2) i have got a value of 0

3: The size of an animal’s pupils expand and contract depending on the amount of light available. Let f(x) = (160x^(-0.4)+90)/(8x^(-0.4)+10)
be the size in mm of the pupils at light intensity x. Find lim(x-->0+) and lim(x-->inf) for f(x), and argue that these represent the largest and smallest possible sizes of the pupils

Can someone please check my answers for 1 and 2 and give me some help for question 3?

P.S sorry about the bad notation for the equation and limits
$\lim_{x\to{0}}\frac{\sin(\sin(x))}{\sin(x)}$

To simplify things a bit we see can see a few things

Firstly you can semi-simplify things by knowing that as $x\to{0}$ $\sin(x)\sim{x}$

which gives $\lim_{x\to{0}}\frac{\sin(\sin(x))}{x}=\lim_{x\to{0 }}\frac{\cos(\sin(x))\cdot{\cos(x)}}{1}=1$

Or more generally $\lim_{x\to{0}}\frac{\sin(u(x))}{u(x)}=1,if\text{ }u(0)=0$

so from that we see that as $x\to{0}$ , $\sin(u(x))\sim{u(x)}$

so we could just say as $x\to{0}$ , $\sin(\sin(x))\sim{\sin(x)}$

so $\lim_{x\to{0}}\frac{\sin(\sin(x))}{\sin(x)}=\lim_{ x\to{0}}\frac{\sin(x)}{\sin(x)}=1$

For the second one apply the squeeze theorem

$\forall{x}>0,\frac{\ln(x)}{x^2}\leq\frac{1}{x}$

and $\forall{x}>0,\frac{1}{x^2}\leq\frac{\ln(x)}{x^2}$

so $\frac{1}{x^2}<\frac{\ln(x)}{x^2}<\frac{1}{x}$

so maintaing equality we introduce the limits

$\lim_{x\to\infty}\frac{1}{x^2}\leq\lim_{x\to\infty }\frac{\ln(x)}{x^2}\leq\lim_{x\to{0}}\frac{1}{x}$

or

$0\leq\lim_{x\to{0}}\frac{\ln(x)}{x^2}\leq{0}$

$\therefore\lim_{x\to{0}}\frac{\ln(x)}{x^2}=0$ by the squeeze theorem

3. Originally Posted by deragon999
1: lim(x-->0) sin(sinx)/sinx...i have solved using l'hopitals rule to get 1

Mr F says: Correct. But no need for the Hospital. Just make the substitution $w = sin x$. Note that as $x \rightarrow 0, ~ w \rightarrow 0$. $~ \lim_{w \rightarrow 0} \frac{\sin w}{w}$ is a standard result.

2: lim(x-->infinity) ln(x)/(x^2) i have got a value of 0

Mr F says: Correct.

3: The size of an animal’s pupils expand and contract depending on the amount of light available. Let f(x) = (160x^(-0.4)+90)/(8x^(-0.4)+10)
be the size in mm of the pupils at light intensity x. Find lim(x-->0+) and lim(x-->inf) for f(x), and argue that these represent the largest and smallest possible sizes of the pupils

Can someone please check my answers for 1 and 2 and give me some help for question 3?

P.S sorry about the bad notation for the equation and limits
3. You have $f(x) = \frac{160 x^{-0.4} + 90}{8 x^{-0.4} + 10} = \frac{160 + 90 x^{0.4}}{8 + 10 x^{0.4}} = 9 + \frac{44}{5 x^{0.4} + 4}$ and the domain is x > 0.

Note that as $x \rightarrow 0^+, x^{0.4} \rightarrow 0$.

Note that as $x \rightarrow +\infty, x^{-0.4} = \frac{1}{x^{0.4}}\rightarrow 0$.

4. Originally Posted by deragon999
1: lim(x-->0) sin(sinx)/sinx...i have solved using l'hopitals rule to get 1

2: lim(x-->infinity) ln(x)/(x^2) i have got a value of 0

3: The size of an animal’s pupils expand and contract depending on the amount of light available. Let f(x) = (160x^(-0.4)+90)/(8x^(-0.4)+10)
be the size in mm of the pupils at light intensity x. Find lim(x-->0+) and lim(x-->inf) for f(x), and argue that these represent the largest and smallest possible sizes of the pupils

Can someone please check my answers for 1 and 2 and give me some help for question 3?

P.S sorry about the bad notation for the equation and limits
For the last one you are asked to find firstly

$\lim_{x\to\infty}\frac{160x^{-.4}+90}{8x^{-.4}+10}=\lim_{x\to\infty}\frac{\frac{160}{x^{.4}}+ 90}{\frac{8}{x^{.4}}+10}
$

Direct substitution yields

$\frac{0+90}{0+10}=9$

And secondly you are asked to find

$\lim_{x\to{0}}\frac{\frac{160}{x^{.4}}+90}{\frac{8 }{x^{.4}}+10}$

Multiplying by $\frac{x^{.4}}{x^{.4}}$

we get

$\lim_{x\to{0}}\frac{160+90x^{.4}}{8+10x^{.4}}$

now direct substitution yields

$\frac{160+0}{8+0}=20$

Now to find if they are the smallest/largest

show that they are the absolute max and min points on

$(0,\infty)$

5. Originally Posted by Mathstud28
and $\forall{x}>0,\frac{1}{x^2}\leq\frac{\ln(x)}{x^2}$

Hmmm....but $x \leq e \Rightarrow \ln x \leq 1$

6. Originally Posted by deragon999
1: lim(x-->0) sin(sinx)/sinx...i have solved using l'hopitals rule to get 1

2: lim(x-->infinity) ln(x)/(x^2) i have got a value of 0

3: The size of an animal’s pupils expand and contract depending on the amount of light available. Let f(x) = (160x^(-0.4)+90)/(8x^(-0.4)+10)
be the size in mm of the pupils at light intensity x. Find lim(x-->0+) and lim(x-->inf) for f(x), and argue that these represent the largest and smallest possible sizes of the pupils

Can someone please check my answers for 1 and 2 and give me some help for question 3?

P.S sorry about the bad notation for the equation and limits
#1: $\text{let} \ u=sin(x)$ and $u(0)=0$. Therefore, the limit is written as $\lim_{u\to{0}}\frac{sin(u)}{u}=\color{red}\boxed{1 }$.

#2 $\lim_{x\to{\infty}}\frac{\ln(x)}{x^2}$.

We get the indeterminate case $\frac{\infty}{\infty}$.

Apply L'Hopital's Rule:

$\lim_{x\to{\infty}}\frac{\ln(x)}{x^2}=\lim_{x\to{\ infty}}\frac{\frac{1}{x}}{2x}=\lim_{x\to{\infty}}\ frac{1}{2x^2}=\color{red}\boxed{0}$

#3 $\lim_{x\to{0^+}}\frac{160x^{-.4}+90}{8x^{-.4}+10}$

If we plug in zero, we get the indeterminate case $\frac{\infty}{\infty}$

Simplifying the limit, we get:

$\lim_{x\to{0^+}}\frac{\frac{160}{x^{.4}}+\frac{90x ^{.4}}{x^{.4}}}{\frac{8}{x^{.4}}+\frac{10x^{.4}}{x ^{.4}}}=\lim_{x\to{0^+}}\frac{\frac{1}{x^{.4}}}{\f rac{1}{x^{.4}}}\cdot\frac{160+90x^{.4}}{8+10x^{.4} }$.

Assuming that $\frac{1}{x^{.4}}\neq 0$, we have:

$\lim_{x\to{0^+}}\frac{160+90x^{.4}}{8+10x^{.4}}$.

Evaluating the limit, we now have: $\frac{160}{8}=\color{red}\boxed{20}$

Now evaluate this as $x\to{\infty}$.

$\lim_{x\to{\infty}}\frac{160+90x^{.4}}{8+10x^{.4}} =\frac{90}{10}=\color{red}\boxed{9}$.

Hope this helped you out!

7. Originally Posted by Isomorphism
Hmmm....but $x \leq e \Rightarrow \ln x \leq 1$
Oh how embarrasing! My calculator apparently cant solve inequalities well!
$\forall{x}>10^{10},\frac{1}{x^2}<\frac{\ln(x)}{x^2 }$

Happy?

8. Originally Posted by Mathstud28
$\lim_{x\to{0}}\frac{\sin(\sin(x))}{\sin(x)}$

To simplify things a bit we see can see a few things

Firstly you can semi-simplify things by knowing that as $x\to{0}$ $\sin(x)\sim{x}$

which gives $\lim_{x\to{0}}\frac{\sin(\sin(x))}{x}=\lim_{x\to{0 }}\frac{\cos(\sin(x))\cdot{\cos(x)}}{1}=1$

Or more generally $\lim_{x\to{0}}\frac{\sin(u(x))}{u(x)}=1,if\text{ }u(0)=0$

so from that we see that as $x\to{0}$ , $\sin(u(x))\sim{u(x)}$

so we could just say as $x\to{0}$ , $\sin(\sin(x))\sim{\sin(x)}$

so $\lim_{x\to{0}}\frac{\sin(\sin(x))}{\sin(x)}=\lim_{ x\to{0}}\frac{\sin(x)}{\sin(x)}=1$
Is it possible to simplify it by making $t = sinx$
$\lim_{x\to{0}}\frac{\sin(t)}{t}$

and solving that?
That's what I did and came up with the answer, 1.

9. for this one:
$\lim_{x\to{0^+}}\frac{160x^{-.4}+90}{8x^{-.4}+10}$
can I apply L'Hopitals rule. Differentiate d/dx:

$\frac{-64x^{-1.4}}{-3.2x^{-1.4}}$

$= \frac{64}{3.2}$

$=20$

Das ist gut, ja?

10. Hello,

Originally Posted by Dr Zoidburg
Is it possible to simplify it by making $t = sinx$
$\lim_{{\color{red}t}\to {0}}\frac{\sin(t)}{t}$

and solving that?
That's what I did and came up with the answer, 1.
Yeah
(tiny mistake, in red)

Originally Posted by Dr Zoidburg
for this one:
$\lim_{x\to{0^+}}\frac{160x^{-.4}+90}{8x^{-.4}+10}$
can I apply L'Hopitals rule. Differentiate d/dx:

$\frac{-64x^{-1.4}}{-3.2x^{-1.4}}$

$= \frac{64}{3.2}$

$=20$

Das ist gut, ja?
Yop