# How to solve these limits ?

1. Jun 2, 2012

1. The problem statement, all variables and given/known data
Find the limits of the following

2. Relevant equations
We can't use L'Hôpital's rule as it's not in our syllabus.We have to use only

3. The attempt at a solution
I attempted a lot but can't find a way to solve these three (out of about 80 limits)
pls show me the steps to get these without L'Hôpital's as Wolfram Alpha give steps in that law...

2. Jun 2, 2012

### Saitama

Mind showing us your attempts first?

Just to give you a start for the first question, can you make it of the form 1?

3. Jun 2, 2012

### Dickfore

The first one is $\infty \cdot 0$, and can be converted to an indeterminate form $\frac{0}{0}$ by representing $\tan x = \frac{1}{\cot x}$. Then, apply L'Hospital's Rule to:
$$\lim_{x \rightarrow \pi/2}{\frac{\ln (\sin x)}{\cot x}}$$
Remember to use chain rule when finding the derivatives.

EDIT: Ah, you can't apply LR. Take $t \equiv \ln(\sin x), x \rightarrow \frac{\pi}{2} \Rightarrow t \rightarrow ?$. Then $\sin x = e^{t}$. Use trigonometry to express $\cot x$.

4. Jun 2, 2012

log(sinx)=log10sinx or log(sinx)=logesinx ?

5. Jun 2, 2012

### Dickfore

Hey, it's your problem. You tell me. The solution changes very little regardless of the base of the lograrithm.

6. Jun 2, 2012

It was given in that way...don't know whether the base is e or 10...
So I googled and found some saying log(x) is normally log10(x) if not given...
So much stuck one this three sums...

7. Jun 2, 2012

### Dickfore

Ok, let's say it's base a. Then $t = \log_{a} (\sin x)$. When $x \rightarrow \pi/2$ we still have $t \rightarrow 0$. When you take the antilogarithm, you have $\sin x = a^{t}$. Now, express cot in terms of sin.

EDIT:
Scratch that!

As post #2 had suggested, notice the following:
$$\tan x \, \log_{a}(\sin x) = \log_{a}{\left[ (\sin x)^{\tan x} \right]} = \log_{a}{\left[ (\sin x)^{\frac{1}{\cot{x}}} \right]}$$

Express all the trigonometric functions in terms of cos (I will tell you why when you do it)!

Last edited: Jun 2, 2012
8. Jun 2, 2012

let's take a=10
cotx=√(cosec2x-1)
cotx=√(10-2t-1)
so we get the limit as ,
lim( t->0){t/√(10-2t-1)}
then how to proceed ?

9. Jun 2, 2012

### Dickfore

see my edit of the post.

10. Jun 2, 2012

sinx=√1-cos2x
secx=1/cosx
cosecx=1/√1-cos2x
tanx=√1-cos2x/cosx
cotx= cosx/√1-cos2x
So what ?

11. Jun 2, 2012

### Dickfore

So, what does the logarithm look like?

12. Jun 2, 2012

No xp with latex sorry....I have no idea brother !
Pls explain !

13. Jun 2, 2012

spends a lot to type Latex...It cost my time and yours

14. Jun 2, 2012

### Dickfore

$$\log_{a} { \left[ (1 - \cos^2 x)^{\frac{\sqrt{1 - \cos^2 x}}{2 \, \cos x}} \right]}$$
Now, take $t = -\cos^2 x \Rightarrow \cos x = \sqrt{-t}$ to get:
$$\log_{a} {\left[ (1 + t)^{\frac{\sqrt{1 + t}}{2 \, \sqrt{-t}}} \right]} = \log_{a}{\left[ (1 + t)^{\frac{1}{t} \, \frac{\sqrt{-t(1 +t)}}{2}}\right]}$$

Do you know the limit
$$\lim_{t \rightarrow 0}(1 + t)^{\frac{1}{t}} = ?$$

If you do, my work here is almost done.

15. Jun 2, 2012

(1+0)1/0=1$\infty$=$\infty$ ?

16. Jun 2, 2012

(1+0)1/0=1$\infty$=$\infty$ ?

17. Jun 2, 2012

### Saitama

No. Aren't you taught any standard limits or basic formulas?

18. Jun 2, 2012

No..only these two are available in our coursework..

Nothing like following in ours
$\lim_{t \rightarrow 0}(1 + t)^{\frac{1}{t}}$

19. Jun 2, 2012

### Saitama

Then, i suggest you to ask your teacher to tell you some more because you will frequently require them.
$$\lim_{x\to 0} (1+x)^{\frac{1}{x}}=e$$
Solving this limit requires use logarithmic properties.

EDIT: Woot! 700 posts.

20. Jun 2, 2012