<p>I am having trouble finding the answers to the following limits. </p>
<p>1) lim [sin(ax)]/[bx]
x-> 0</p>
<p>2) lim [sin(ax)]/[sin(bx)]
x-> 0</p>
<p>3) lim [tan(ax)]/[bx]
x-> 0</p>
<p>4) lim [tan(ax)]/[tan(bx)]
x-> 0</p>
<p>5) lim [sin(ax)]/[tan(bx)]
x-> 0</p>
<p>Basically all the questions want you to find the limit as x approaches zero for fractions. In the first problem the numerator is sin(ax) and the denominator is bx. In the second problem the numerator is sin(ax) and the denominator is sin(bx). In the third problem the numerator is tan(ax) and the denominator is bx. In the fourth problem the numerator is tan(ax) and the denominator is tan(bx). In the fifth problem the numerator is sin(ax) and the denominator is tan(bx).</p>
<p>1) a/b
The important part of this question is small angle approximation for sin(x). For x<<1, sin(x) = x. So,
lim sin(ax)/(bx) = lim ax/bx = a/b
x-> 0 x->0
You could also solve this problem with L'hopital's rule.
2) a/b
For this one, again, remember that as x<<1, sin(x) = x so,
lim sin(ax)/sin(bx) = lim ax/bx = a/b
x-> 0 x->0
3) a/b
See the justification for 1 and 2, tan(x) approaches x just as sin(x) does.
4) a/b
Again, the same.
5) a/b
Finally, a combination using both small angle approximations.</p>
<p>can you use L' Hopitals rule for all the five problems? because we didn't learn angle approximations in my class.</p>
<p>There is a much simpler way of solving the first problem. You know that as x approaches zero that the limit of sina/a is one. So mulptiply the whole fraction by a/a and you get sinax/ax * a/b. The first fraction goes to one so the answer is a/b.</p>
<p>I still don't understand</p>
<p>I think using L' Hopitals rule is the best way to solve this problem for me.</p>