# Limit problem

#### dome123

##### New member
Hi,
I have a problem which needs to be solved without using L'Hosiptal rule.
Can anyone help me?
Thanks.

#### firemath

##### Full Member
Ok. I'm trying to graph it....

#### firemath

##### Full Member
Oh dear.

Let me try this a different way.

#### dome123

##### New member
Is it possible to do a substitution where cosx=t?

#### firemath

##### Full Member
Ok. I think this is right, but this is not my forte.

So first, convert both sqrt of cos x and the cubic root of cos x into exponent form.
Then, convert both of their powers to be the same.

See what you can do with that.

#### dome123

##### New member
What do you mean exactly by converting their powers to be the same, I can't see how to do it.

#### pka

##### Elite Member
This is a truly awful question. Even the most powerful computer algebra system takes pages to solve this and the even for the last few steps uses L'Hosiptal rule. I see no straightforward way of doing it. WolframAlpha gives the answer as $$\displaystyle -\frac{1}{12}$$.
SEE HERE one needs to Pro Premium edition to get the steps.

#### dome123

##### New member
Well, pka thanks for your reply , I know it's not easy but this is the question on my exam and I was struggling to solve it but now I see it is too hard.
Thanks anyway

#### Subhotosh Khan

##### Super Moderator
Staff member
Hi,
I have a problem which needs to be solved without using L'Hosiptal rule.
Can anyone help me?
Thanks.View attachment 15947

Please follow the rules of posting in this forum, as enunciated at:

Hint:

If you factor out $$\displaystyle \sqrt[3]{cos(x)}$$ from the numerator - what would you get?

#### Dr.Peterson

##### Elite Member
Well, pka thanks for your reply , I know it's not easy but this is the question on my exam and I was struggling to solve it but now I see it is too hard.
Thanks anyway
No, it isn't too hard. Computers are not always smarter than people.

The suggests that have been given allowed me to work it out in about two square inches of paper. (Well, I write very small -- that accounts for some of my errors.) Your substitution t = cos(x) does make things a lot easier to write, factoring out t^(1/3) helps more, and some conjugates and the like finish the job. Don't give up!

#### Jomo

##### Elite Member
Is it possible to do a substitution where cosx=t?
You can make any substitution you like. How did this one go? Remember if cos(x)=t, then sin^2(x)= 1 - t^2

#### dome123

##### New member
I've solved it by using the supstitution cosx=t^6.
Thank you all for taking your time to reply and trying to help.

#### JeffM

##### Elite Member
NICE JOB. Your substitution gets rid of all the surds. Assuming 0 < cos(x) < 1

$$\displaystyle cos(x) = t^6 \implies \sqrt{cos(x)} = t^3,\ \sqrt[3]{cos(x)} = t^2,\ cos^2(x) = t^{12},$$

$$\displaystyle sin^2(x) = 1 - cos^2(x) = 1 - t^{12} = (1 - t^3)(1 + t^3)(1 + t^6) = (1 - t)(1 + t + t^2)(1 + t^3)(1 + t^6) \text { and}$$

$$\displaystyle t^3 - t^2 = t^2((t = 1) = -\ t^2(1 - t).$$

$$\displaystyle \therefore y = \dfrac{\sqrt{cos(x)} - \sqrt[3]{cos(x)}}{sin^2(x)} = \dfrac{t^3 - t^2}{1 - t^{12}} =$$

$$\displaystyle \dfrac{-\ t^2 \cancel {(1 - t)}}{\cancel {(1 - t)}(1 + t + t^2)(1 + t^3)(1 + t^6)} = -\ \dfrac{t^2}{(1 + t + t^2)(1 + t^3)(1 + t^6)}.$$

$$\displaystyle \lim_{x \rightarrow 0} cos(x) = 1 \implies \lim_{x \rightarrow 0} t^6 = 1 \implies \lim_{x \rightarrow 0} t^n =1 \implies$$

$$\displaystyle \lim_{x \rightarrow 0} y = -\ \dfrac{1^2}{(1 + 1^1 + 1^2)(1 + 1^3)(1 + 1^6)} = -\ \dfrac{1}{3(2)(2)} = -\ \dfrac{1}{12}.$$

If you ask a machine just to give you the answer, the reason for the answer will not be apparent.

#### dome123

##### New member
Exactly, at first glance I was wondering how to factor it but I realised I forgot the basics, thanks for the reply.