# How to solve these limits ?

## Homework Statement

Find the limits of the following

## Homework Equations

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

## 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...

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Mind showing us your attempts first?

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

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 $\cot x$.

can you make it of the form 1?

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 $\cot x$.
log(sinx)=log10sinx or log(sinx)=logesinx ?

log(sinx)=log10sinx or log(sinx)=logesinx ?
Hey, it's your problem. You tell me. The solution changes very little regardless of the base of the lograrithm.

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

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)!

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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
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 ?

see my edit of the post.

$\tan x \, \log_{a}(\sin x) = \log_{a}{\left[ (\sin x)^{\tan x} \right]} = \log_{a}{\left[ (\sin x)^{\frac{1}{\cot{x}}} \right]}$

Express all the trigonometric functions in terms of cos (I will tell you why when you do it)!
sinx=√1-cos2x
secx=1/cosx
cosecx=1/√1-cos2x
tanx=√1-cos2x/cosx
cotx= cosx/√1-cos2x
So what ?

So, what does the logarithm look like?

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

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

$$\log_{a} { \left[ (1 - \cos^2 x)^{\frac{\sqrt{1 - \cos^2 x}}{2 \, \cos x}} \right]}$$
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.

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

If you do, my work here is almost done.
(1+0)1/0=1$\infty$=$\infty$ ?

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

(1+0)1/0=1$\infty$=$\infty$ ?
No. Aren't you taught any standard limits or basic formulas?

No. Aren't you taught any standard limits or basic formulas?
No..only these two are available in our coursework..

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

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

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

Then, i suggest you to ask your teacher to tell you some more because you will frequently require them.
That's what I'm gong to do next..
Anyway what about the other two limits ?
[2] and [3] ?

EDIT: Woot! 700 posts.
congratz !

EDIT: Woot! 37 posts.

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What is $\frac{1}{\sqrt{2}}$? Can you write it in a trigonometric form?
And expand cot(x) as cos(x)/sin(x).

For (2), since x->a, which is unknown constant, the limit does not give you any indeterminate messy form, so all you need to do is substitute x=a.

For (2), since x->a, which is unknown constant, the limit does not give you any indeterminate messy form, so all you need to do is substitute x=a.
Oh yes, that's the thing i came across in [2] .
@shalikadm: Is the [2] question correct? Just check it once again from where you got it.

For (2), since x->a, which is unknown constant, the limit does not give you any indeterminate messy form, so all you need to do is substitute x=a.
Oh yes, that's the thing i came across in [2] .
@shalikadm: Is the [2] question correct? Just check it once again from where you got it.
OMG..sorry...it's not x to a ,but x to 0....sorry