Determine the following limit - but do not use L'Hopital's rule:
\(\lim_{x\rightarrow 9}\frac{sin(\sqrt{x}-3)}{x-9}\) Hint: Let \(t =\sqrt{x}\)
I would really appreciate some assistance on this. I know that as x-> 9, t->3, but I don't know how to adjust my limit to suit this. Thanks
Let's use the hint, \(\text{If} \space \space t=\sqrt{x} \iff t^2=x\)
We get: \(lim_{t->3}\dfrac{sin(t-3)}{t^2-9}\)
Factor the denominator:
\(lim_{t->3}\dfrac{sin(t-3)}{(t-3)(t+3)}\)
Then, let's rewrite this as:
\(lim_{t->3}[\dfrac{sin(t-3)}{(t-3)}*\dfrac{1}{t+3}]\)
Using the following law of limits: \(lim_{x->a}[g(x)f(x)]=(lim_{x->a}g(x))*(lim_{x->a}f(x))\)
So we get:
\(lim_{t->3}[\dfrac{sin(t-3)}{(t-3)}*\dfrac{1}{t+3}]=[lim_{t->3}\dfrac{sin(t-3)}{(t-3)}]*[lim_{t->3}\dfrac{1}{t+3}]\)
Recall: \(lim_{y->0}\frac{sin(y)}{y}=1\) (You could find the proof of this on youtube, but this is usually a standard result; that is, when you are solving a question, you do not need to prove this.)
Now, (t-3) as t-->3 will go to 0 and (t-3) as t goes to 3 will go to 0 as well, so basically this is the same as: \(lim_{y->0}\frac{sin(y)}{y}=1\)
(Where y=t-3). Hence, the limit in the first bracket is just 1, and the second limit in the other bracket could be done by direct substitution.
So:
\(1*(\dfrac{1}{3+3})=\dfrac{1}{6}\) which is our answer! :)
I hope this helps!
