lim[x->0](tanbx/sinbx)
lim[x->0]{(sinbx/cosbx)/sinbx} x cosbx/cosbx
This is almost right. What you said is:
<br />
\lim_{x \rightarrow 0} \frac{\tan bx}{\sin bx}<br />
= \left( \lim_{x \rightarrow 0} \frac{ \frac{\sin bx}{\cos bx} }{ \sin bx }<br />
\right) \times \frac{\cos bx}{\cos bx}<br />
However, this is wrong: it only makes sense to use x
inside the limit, not outside. What you wanted to say was:
<br />
\lim_{x \rightarrow 0} \frac{\tan bx}{\sin bx}<br />
= \lim_{x \rightarrow 0} \left( \frac{ \frac{\sin bx}{\cos bx} }{ \sin bx }<br />
\times \frac{\cos bx}{\cos bx} \right)<br />
So anyways, in the end you're left with \lim_{x \rightarrow 0} 1 / \cos bx... why do you think you haven't gotten anywhere?
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The second problem is entirely wrong. Some things to note are:
The theorem
<br />
\lim_{x \rightarrow a} \frac{f(x)}{g(x)}<br />
= \frac{ \lim_{x \rightarrow a} f(x) }{ \lim_{x \rightarrow a} g(x) }<br />
is true only when both of the individual limits exist
and the denominator of the right hand side is not zero.
And each of these following statements are usually
wrong:
\sin^n x = \sin x^n
\sin (xy) = (\sin x) \times y
In the interest of saving time, I'll remind you that \lim_{x \rightarrow 0} \sin x / x = 1.
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For the third problem, your work looks right, however your conclusion is probably slightly wrong. In particular, your teacher probably wants you to say that g(x) is undefined at x = -3, but g(x) = x - 2 everywhere else.
