Help! Lim: y = l*cos(ax)/(ax^2), x < 0; 13/2, x = 0; 1n(1+ax)/(bx), x > 0

[gilad]

help me please:

\(\displaystyle y\, =\, \begin{cases} \dfrac{l\, \cdot\, \cos(a\, \cdot\, x)}{a\, \cdot\, x^2},&\mbox{ for }\, x\, <\, 0 \\ \dfrac{13}{2},&\mbox{ for }\, x\, =\, 0 \\ \dfrac{1n(1\, +\, a\, \cdot\, x)}{b\, \cdot\, x},&\mbox{ for }\, x\, >\, 0 \end{cases}\)

For which a, b values function continuous in x=0 and y=(picture)?

 

[Deleted member 4993]

help me pleas. For which a,b values function continuous in x=0 and y=(picture)?

What are your thoughts?

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[stapel]

help me please:

We'll be glad to! But first we'll need to see what you've tried and how far you've gotten, so we know what kind of help you're needing.

For which values of a and b will the function y (below) be continuous at x = 0?

I am guessing that the instructions are meant to say something like the above.

\(\displaystyle y\, =\, \begin{cases} \dfrac{l\, \cdot\, \cos(a\, \cdot\, x)}{a\, \cdot\, x^2},&\mbox{ for }\, x\, <\, 0 \\ \dfrac{13}{2},&\mbox{ for }\, x\, =\, 0 \\ \dfrac{1n(1\, +\, a\, \cdot\, x)}{b\, \cdot\, x},&\mbox{ for }\, x\, >\, 0 \end{cases}\)

For some reason, some of the characters seem to be mixed up. For instance, in the "x < 0" portion, there is no need for a "1" in front of the cosine, since multiplying by 1 doesn't change anything. In fact, it appears to be a variable "ell". But what information did they give you about this variable?

For the "x > 0" portion, the natural log, rather than being written "ln" as usual, appears to be written as "1n". Am I mistaken about this being a log? Is this really another extraneous "1"? If so, then what information are you given about the variable "n"?

Thank you!

 

[Ishuda]

help me please:

\(\displaystyle y\, =\, \begin{cases} \dfrac{l\, \cdot\, \cos(a\, \cdot\, x)}{a\, \cdot\, x^2},&\mbox{ for }\, x\, <\, 0 \\ \dfrac{13}{2},&\mbox{ for }\, x\, =\, 0 \\ \dfrac{1n(1\, +\, a\, \cdot\, x)}{b\, \cdot\, x},&\mbox{ for }\, x\, >\, 0 \end{cases}\)

For which a, b values function continuous in x=0 and y=(picture)?

Isn't that
\(\displaystyle y\, =\, \dfrac{1\, -\, \cos(a\, \cdot\, x)}{a\, \cdot\, x^2},\, \mbox{ for }\, x\, <\, 0\)?

 

[stapel]

Isn't that
\(\displaystyle y\, =\, \dfrac{1\, -\, \cos(a\, \cdot\, x)}{a\, \cdot\, x^2},\, \mbox{ for }\, x\, <\, 0\)?

If only the original poster would reply....