How to solve these limits ?

  • Thread starter shalikadm
  • Start date
  • #1
63
0

Homework Statement


Find the limits of the following
148lrc.png


Homework Equations


We can't use L'Hôpital's rule as it's not in our syllabus.We have to use only
263d79g.jpg


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...
 

Answers and Replies

  • #2
3,812
92
Mind showing us your attempts first?
Nobody will help you if you don't show your efforts.

Just to give you a start for the first question, can you make it of the form 1?
 
  • #3
2,967
5
The first one is [itex]\infty \cdot 0[/itex], and can be converted to an indeterminate form [itex]\frac{0}{0}[/itex] by representing [itex]\tan x = \frac{1}{\cot x}[/itex]. Then, apply L'Hospital's Rule to:
[tex]
\lim_{x \rightarrow \pi/2}{\frac{\ln (\sin x)}{\cot x}}
[/tex]
Remember to use chain rule when finding the derivatives.

EDIT: Ah, you can't apply LR. Take [itex]t \equiv \ln(\sin x), x \rightarrow \frac{\pi}{2} \Rightarrow t \rightarrow ?[/itex]. Then [itex]\sin x = e^{t}[/itex]. Use trigonometry to express [itex]\cot x[/itex].
 
  • #4
63
0
can you make it of the form 1?
No idea about 1 form..

EDIT: Ah, you can't apply LR. Take [itex]t \equiv \ln(\sin x), x \rightarrow \frac{\pi}{2} \Rightarrow t \rightarrow ?[/itex]. Then [itex]\sin x = e^{t}[/itex]. Use trigonometry to express [itex]\cot x[/itex].
log(sinx)=log10sinx or log(sinx)=logesinx ?
 
  • #5
2,967
5
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.
 
  • #6
63
0
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...
 
  • #7
2,967
5
Ok, let's say it's base a. Then [itex]t = \log_{a} (\sin x)[/itex]. When [itex]x \rightarrow \pi/2[/itex] we still have [itex]t \rightarrow 0[/itex]. When you take the antilogarithm, you have [itex]\sin x = a^{t}[/itex]. Now, express cot in terms of sin.

EDIT:
Scratch that!

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

Express all the trigonometric functions in terms of cos (I will tell you why when you do it)!
 
Last edited:
  • #8
63
0
Ok, let's say it's base a. Then [itex]t = \log_{a} (\sin x)[/itex]. When [itex]x \rightarrow \pi/2[/itex] we still have [itex]t \rightarrow 0[/itex]. When you take the antilogarithm, you have [itex]\sin x = a^{t}[/itex].
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 ?
 
  • #9
2,967
5
see my edit of the post.
 
  • #10
63
0
[itex]
\tan x \, \log_{a}(\sin x) = \log_{a}{\left[ (\sin x)^{\tan x} \right]} = \log_{a}{\left[ (\sin x)^{\frac{1}{\cot{x}}} \right]}
[/itex]

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 ?
 
  • #11
2,967
5
So, what does the logarithm look like?
 
  • #12
63
0
No xp with latex sorry....I have no idea brother !
Pls explain !
 
  • #13
63
0
spends a lot to type Latex...It cost my time and yours
 
  • #14
2,967
5
[tex]
\log_{a} { \left[ (1 - \cos^2 x)^{\frac{\sqrt{1 - \cos^2 x}}{2 \, \cos x}} \right]}
[/tex]
Now, take [itex]t = -\cos^2 x \Rightarrow \cos x = \sqrt{-t}[/itex] to get:
[tex]
\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]}
[/tex]

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

If you do, my work here is almost done.
 
  • #15
63
0
Do you know the limit
[tex]
\lim_{t \rightarrow 0}(1 + t)^{\frac{1}{t}} =?
[/tex]

If you do, my work here is almost done.
(1+0)1/0=1[itex]\infty[/itex]=[itex]\infty[/itex] ?
 
  • #16
63
0
(1+0)1/0=1[itex]\infty[/itex]=[itex]\infty[/itex] ?
 
  • #17
3,812
92
(1+0)1/0=1[itex]\infty[/itex]=[itex]\infty[/itex] ?
No. Aren't you taught any standard limits or basic formulas?
 
  • #18
63
0
No. Aren't you taught any standard limits or basic formulas?
No..only these two are available in our coursework..
263d79g.jpg

Nothing like following in ours
[itex]
\lim_{t \rightarrow 0}(1 + t)^{\frac{1}{t}}
[/itex]
 
  • #19
3,812
92
No..only these two are available in our coursework..
263d79g.jpg

Nothing like following in ours
[itex]
\lim_{t \rightarrow 0}(1 + t)^{\frac{1}{t}}
[/itex]
Then, i suggest you to ask your teacher to tell you some more because you will frequently require them.
[tex]\lim_{x\to 0} (1+x)^{\frac{1}{x}}=e[/tex]
Solving this limit requires use logarithmic properties.

EDIT: Woot! 700 posts.
 
  • #20
63
0
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] ?
 
  • #21
63
0
Last edited:
  • #22
3,812
92
I don't know about [2] but i can help you out with [3].

What is [itex]\frac{1}{\sqrt{2}}[/itex]? Can you write it in a trigonometric form?
And expand cot(x) as cos(x)/sin(x).
 
  • #23
881
40
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.
 
  • #24
3,812
92
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.
 
  • #25
63
0
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
 

Related Threads on How to solve these limits ?

  • Last Post
Replies
10
Views
883
Replies
3
Views
3K
Replies
3
Views
879
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
2
Views
2K
Replies
1
Views
1K
  • Last Post
Replies
13
Views
2K
Replies
4
Views
1K
Top