Limit Solving Strategies for Non-L'Hôpital's Rule Problems

[shalikadm]

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...

 

[Saitama]

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^∞?

 

[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:
<br /> \lim_{x \rightarrow \pi/2}{\frac{\ln (\sin x)}{\cot x}}<br />
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.

 

[shalikadm]

can you make it of the form 1^∞?

No idea about 1^∞ form..

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)=log_esinx ?

 

[Dickfore]

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

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

 

[shalikadm]

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...

 

[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:
<br /> \tan x \, \log_{a}(\sin x) = \log_{a}{\left[ (\sin x)^{\tan x} \right]} = \log_{a}{\left[ (\sin x)^{\frac{1}{\cot{x}}} \right]}<br />

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

 

[shalikadm]

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 ?

 

[Dickfore]

see my edit of the post.

 

[shalikadm]

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

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 ?

 

[Dickfore]

So, what does the logarithm look like?

 

[shalikadm]

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

 

[shalikadm]

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

 

[Dickfore]

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

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

If you do, my work here is almost done.

 

[shalikadm]

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

If you do, my work here is almost done.

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

 

[shalikadm]

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

 

[Saitama]

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

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

 

[shalikadm]

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
<br /> \lim_{t \rightarrow 0}(1 + t)^{\frac{1}{t}}<br />

 

[Saitama]

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

Nothing like following in ours
<br /> \lim_{t \rightarrow 0}(1 + t)^{\frac{1}{t}}<br />

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.

 

[shalikadm]

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] ?

 

[shalikadm]

EDIT: Woot! 700 posts.

congratz !

EDIT: Woot! 37 posts.

 

[Saitama]

I don't know about [2] but i can help you out with [3].

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

 

[Infinitum]

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.

 

[Saitama]

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.

 

[shalikadm]

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

 

[shalikadm]

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

\frac{1}{\sqrt{2}} as cos π/4 or sin π/4 ?

 

[Infinitum]

OMG..sorry...it's not x to a ,but x to 0...sorry

Do you know the definition of derivative of a function in limit form? Question (2) is the simple definition of derivative of f(y) in that form. All you need to do is find what f(y) is and then find f'(y).

 

[shalikadm]

Do you know the definition of derivative of a function in limit form? Question (2) is the simple definition of derivative of f(y) in that form. All you need to do is find what f(y) is and then find f'(y).

urgh..then its the derivative of ysec(y)...so then ?
Is it the answer ?-ysec(y) ?

 

[Infinitum]

urgh..then its the derivative of ysec(y)...so then ?
Is it the answer ?-ysec(y) ?

Yes, the answer is the derivative of y*sec(y). But you need to apply product rule while differentiating this function!

 

[shalikadm]

Yes, the answer is the derivative of y*sec(y). But you need to apply product rule while differentiating this function!

We have not such rule in Limits but in differential calculus we were taught,
y=f(x)g(x)
\frac{dy}{dx}=g(x)f'(x)+f(x)g'(x)

 

[Infinitum]

We have not such rule in Limits but in differential calculus we were taught,
y=f(x)g(x)
\frac{dy}{dx}=g(x)f'(x)+f(x)g'(x)

Exactly. Use this rule to differentiate y*sec(y), that's the answer you are looking for!

 

[shalikadm]

what about the [3] one?

 

[Infinitum]

what about the [3] one?

Convert \frac{1}{\sqrt{2}} into cos(\frac{\pi}{4}), and apply the formula that changes the difference of two trigonometric terms into their product. In the denominator, after writing cot in terms of cos and sin, try getting a single trigonometric ratio and simplify.

 

[shalikadm]

thanks Infinitum and Pranav-Arora...Successfully solved the 2nd and 3rd limits..I'm going to ask about 1st one from a teacher...thanks for the help !