a few limit probs: lim x->-infty sqrtx-4/3x+5, x->pi/2 1-sin

[Opus89]

Hi everyone,
I'm new here and I need a little help. I am currently in Calculus 1 in college, and while math has never been my strong point, I have to take classes like these for my major. Anyway here are the problems:
1.) lim x-> -? ? x-4 / 3x+5 I tried to graph this one on a ti-83 and i couldnt get a graph to show up.

2.) lim ?x->0 (1/?x+?x - 1/?x)/?x

3.) lim x-> ?/2 1-sinx/cosx I tried to rationalize the numerator on this one and still wound up with a zero on the bottom.

I don't expect you guys to tell me the answers. I'm just looking for some guidance and a push in the right direction. Thanks

 

[Loren]

Re: a few limit problems

I think that the first thing you need to learn is how to indicate what is meant. You need to concentrate on using grouping symbols to accurately portray what is meant. I'll use your first problem as an example.

? x-4 / 3x+5 means \(\displaystyle \sqrt{x} - \frac{4}{3x} + 5\).

I don't think that's what is meant. But I don't know what to guess. Is it

?( x-4) /( 3x+5) which means \(\displaystyle \frac{\sqrt{x-4}}{3x+5}\)

or

?(( x-4) /( 3x+5)) which means \(\displaystyle \sqrt{\frac{x-4}{3x+5}}\)

or something else???

 

[skeeter]

Re: a few limit problems

Hi everyone,
I'm new here and I need a little help. I am currently in Calculus 1 in college, and while math has never been my strong point, I have to take classes like these for my major. Anyway here are the problems:
1.) lim x-> -? ? x-4 / 3x+5 I tried to graph this one on a ti-83 and i couldnt get a graph to show up.

that's probably because the function is undefined for negative values of x. recheck ... should it be x -> infinity?
also, is it ?x - 4 or ?(x - 4) and 3x+5 or (3x+5)? use of grouping symbols to enhance the clarity of your mathematical expressions is encouraged here.

2.) lim ?x->0 (1/?x+?x - 1/?x)/?x

rationalize the numerator by multiplying by [1/?(x+?x) + 1/?x]/[1/?(x+?x) + 1/?x], then add the fractions in the numerator and simplify.

3.) lim x-> ?/2 1-sinx/cosx I tried to rationalize the numerator on this one and still wound up with a zero on the bottom.

how did you "rationalize" the numerator? did you multiply by (1+sinx)/(1+sinx) ? if so, show what you did.

I don't expect you guys to tell me the answers. I'm just looking for some guidance and a push in the right direction. Thanks

 

[Opus89]

Re: a few limit problems

I think that the first thing you need to learn is how to indicate what is meant. You need to concentrate on using grouping symbols to accurately portray what is meant. I'll use your first problem as an example.

? x-4 / 3x+5 means \(\displaystyle \sqrt{x} - \frac{4}{3x} + 5\).

I don't think that's what is meant. But I don't know what to guess. Is it

?( x-4) /( 3x+5) which means \(\displaystyle \frac{\sqrt{x-4}}{3x+5}\) <----------- It's this one

or

?(( x-4) /( 3x+5)) which means \(\displaystyle \sqrt{\frac{x-4}{3x+5}}\)

or something else???

"how did you "rationalize" the numerator? did you multiply by (1+sinx)/(1+sinx) ? if so, show what you did."
I multiplied the top and the bottom by 1+sinx the conjugate of the numerator(1-sinx).

Also, the first problem has x going to negative infinity. Maybe this is grouped a little better. ?(x - 4) / (3x+5) I got something to show up on the graph, I had to change some things in the zoom.

 

[skeeter]

Re: a few limit problems

"how did you "rationalize" the numerator? did you multiply by (1+sinx)/(1+sinx) ? if so, show what you did."
I multiplied the top and the bottom by 1+sinx the conjugate of the numerator(1-sinx).

\(\displaystyle \frac{1-\sin{x}}{\cos{x}} \cdot \frac{1+\sin{x}}{1+\sin{x}} = \frac{1-\sin^2{x}}{\cos{x}(1 + \sin{x})} = \frac{\cos^2{x}}{\cos{x}(1+\sin{x})} = \frac{\cos{x}}{1+\sin{x}}\)

now, how did you get 0 in the denominator?

Also, the first problem has x going to negative infinity. Maybe this is grouped a little better. ?(x - 4) / (3x+5) I got something to show up on the graph, I had to change some things in the zoom.

as stated before, \(\displaystyle \frac{\sqrt{x-4}}{3x+5}\) is undefined for values of x < 4 ... ???

 

[Opus89]

Re: a few limit problems

"how did you "rationalize" the numerator? did you multiply by (1+sinx)/(1+sinx) ? if so, show what you did."
I multiplied the top and the bottom by 1+sinx the conjugate of the numerator(1-sinx).

\(\displaystyle \frac{1-\sin{x}}{\cos{x}} \cdot \frac{1+\sin{x}}{1+\sin{x}} = \frac{1-\sin^2{x}}{\cos{x}(1 + \sin{x})} = \frac{\cos^2{x}}{\cos{x}(1+\sin{x})} = \frac{\cos{x}}{1+\sin{x}}\)

now, how did you get 0 in the denominator?

[quote:128d51vg]Also, the first problem has x going to negative infinity. Maybe this is grouped a little better. ?(x - 4) / (3x+5) I got something to show up on the graph, I had to change some things in the zoom.

as stated before, \(\displaystyle \frac{\sqrt{x-4}}{3x+5}\) is undefined for values of x < 4 ... ???[/quote:128d51vg]

Ok i think i see. If I remember correctly, isn't there some "theorem" that says something like 1-sin^2x = cos^2 x? If so, i think i see where i was messing up.