Limits Help... work shown... please help

J

johnq2k7

New member
Joined
Feb 10, 2009
Messages
46
Question:

a.) lim x-->infinity {(3x^3)+ cos(x)} / {(sin(x)- x^3)}

b.) lim x-->pi(+) {(tan^-1 (1/(x-pi))} / {pi-x0}

c.) lim x--> 0(+) {sqrt(x+ sin(x))}*(ln x)

d.) lim x--> 1(-) {(cos^-1 (x))} / (1-x)


My work process:

a.) I divided the numerator and the denominator by the highest power which is 3... therefore
the expression becomes {3+ cos(x)/3}/ {sin (x)/3 - 1)}
since, 1/infinity becomes zero
the limit should be -3

b.)for the numerator of the expression as x approaches pi from the positive side the inverse of tangent is 1/0 which is undefined... so the expression of the numerator is undefined..... also... as x approaches pi from the positive side the denominator of the expression approaches 0.. so the overall expression becomes 1/0... therefore the limit does not exist.. and since it is not bound... i'm sort of iffy on my answer for this one in terms of my reasoning and evaluation please help me out

c.) as x approaches zero from the positive side, the square root of the expression x+ sin(x) becomes zero, however the natural logarithm of zero is undefined... therefore the product of 0 and an undefined quantity doesn't produce a limit... therefore the limit is undefined... i'm sort of iffy on my answer for this question as well


d.) as approaches 1 from the negative side the numerator of the expression of inverse cosine of x becomes zero, the denominator also becomes zero since the expression of the denominator is 1-x.... therefore the limit is zero for this expression.. i'm sort of iffy for this expression as well


Please help me with these problems... i have provided some of my work process
 
johnq2k7 said:
Question:

a.) lim x-->infinity {(3x^3)+ cos(x)} / {(sin(x)- x^3)}

b.) lim x-->pi(+) {(tan^-1 (1/(x-pi))} / {pi-x0}

c.) lim x--> 0(+) {sqrt(x+ sin(x))}*(ln x)

d.) lim x--> 1(-) {(cos^-1 (x))} / (1-x)


My work process:

a.) almost correct...
I divided the numerator and the denominator by the highest power which is 3... therefore
the expression becomes

{3+ cos(x)/x^3}/ {sin (x)/x^3 - 1)}
since, 1/infinity becomes zero
the limit should be -3 <<<< Correct

b.)for the numerator of the expression as x approaches pi from the positive side the inverse of tangent is 1/0 which is undefined... so the expression of the numerator is undefined..... also... as x approaches pi from the positive side the denominator of the expression approaches 0.. so the overall expression becomes 1/0... therefore the limit does not exist.. and since it is not bound... i'm sort of iffy on my answer for this one in terms of my reasoning and evaluation please help me out

c.) as x approaches zero from the positive side, the square root of the expression x+ sin(x) becomes zero, however the natural logarithm of zero is undefined... therefore the product of 0 and an undefined quantity doesn't produce a limit... therefore the limit is undefined... i'm sort of iffy on my answer for this question as well


d.) as approaches 1 from the negative side the numerator of the expression of inverse cosine of x becomes zero, the denominator also becomes zero since the expression of the denominator is 1-x.... therefore the limit is zero for this expression.. i'm sort of iffy for this expression as well


Please help me with these problems... i have provided some of my work process
Click to expand...

Your reasoning seems okay to me.

Using a graphing calculator - plot the function to reinforce your reasons.
 
When inputted answers for b.) I got negative infinity as an answer instead of limit does not exist.... i'm confused with that one


for c.) i got positive infinity from the calculator instead of limit doesn't exist... i'm confused


d.) i got postive infinity instead of zero from the calculator i'm confused for that one as well
 
For part c.

You have \(\displaystyle \lim_{x\to 0^{+}}\sqrt{x+sin(x)}\cdot ln(x)\).

Have you covered L'Hopital's rule yet?. That may be handy.

If we apply the rule several times, we get \(\displaystyle -3\sqrt{\lim_{x\to 0^{+}}sin(x)}=0\)

Graph this and you can see as x approaches 0 from the right the limit is 0, not positive infinity.

part d. \(\displaystyle \lim_{x\to 1^{-}}\frac{cos^{-1}(x)}{1-x}\). The limit is unbounded.

By applying L'Hopital, we get:

\(\displaystyle \lim_{x\to 1^{-}}\frac{1}{\sqrt{1-x^{2}}}=\infty\)

It is unbounded and does not exist.
 
Thanks for your help... so far I understand my answers for a,c,d ... however is my answer for b.) correct?

here is my answer again:


b.)for the numerator of the expression as x approaches pi from the positive side the inverse of tangent is 1/0 which is undefined... so the expression of the numerator is undefined..... also... as x approaches pi from the positive side the denominator of the expression approaches 0.. so the overall expression becomes 1/0... therefore the limit does not exist.. and since it is not bound... i'm sort of iffy on my answer for this one in terms of my reasoning and evaluation please help me out
 
If show a sketch of the curve near x = ?, then you would not need all those description.

In general, I gather, you are saying as x ? ?, f(x) ? undefined and that is correct.
 
when I computed the limit for b.) using a graphing calculator I got negative infinity as an answer instead of limit does not exist which is confusing.... however I had expected to get limit does not exist or is undefined.... did the graphing calculator compute the value as negative infinity since the x value is approaching a number from the positive end and it isn't a two sided limit.. therefore instead of undefined or limit does not exist the graphing calculator computes negative infinity as a value?
 
You must log in or register to reply here.
Top