Solve Limit approaches 0 from both sides (sqrt15-6(1))+h)-(sqrt15-6(1))?
... How do I begin? This was originally a Newton's Difference Quotient but i've already plugged the numbers into the formula and what not and this is what I got and the square roots are confusing me! Can somebody help me?
Please read Read before Posting before you post again. It is always important that we understand what is your general level of math education (and usually what course you are currently taking) so we can write an answer that makes sense to
YOU.
It is also important that you quote your problem exactly and completely so that we are certain what you are asking about. You say this is about a quotient, but there is no division shown or even implied in what you gave us above. We are forced to guess.
I am going to guess that the word quotient means that what you are really trying to solve is something like this:
\(\displaystyle Given\ x < \dfrac{5}{2}\ and\ f(x) =\sqrt{15 - 6x},\ find\ \displaystyle \lim_{h \rightarrow 0}\dfrac{f(x + h) - f(x)}{h}.\)
If that is the case, you are on the wrong track. An important law of limits is
\(\displaystyle a,\ b \in \mathbb R\ and \displaystyle \lim_{x \rightarrow c}p(x) = a\ and\ \lim_{x \rightarrow c}q(x) = b \ne 0 \implies \lim_{x \rightarrow c}\dfrac{p(x)}{q(x)} = \dfrac{a}{b}.\)
Notice the condition that b
NOT EQUAL zero. Pretty clearly \(\displaystyle \displaystyle \lim_{h \rightarrow 0}h = 0.\)
So it will do you no good to evaluate the limit of f(x + h) - f(x) as h approaches 0 because the law of limits that you are trying to use does not apply.
Here is an important trick about square roots to learn
\(\displaystyle r(x) = \sqrt{s(x)}\ and\ h \ne 0 \implies \dfrac{r(x + h) - r(x)}{h} = \dfrac{\sqrt{s(x + h)} - \sqrt{s(x)}}{h} \implies\)
\(\displaystyle \dfrac{r(x + h) - r(x)}{h} = \dfrac{\sqrt{s(x + h)} - \sqrt{s(x)}}{h} * 1 = \dfrac{\sqrt{s(x + h)} - \sqrt{s(x)}}{h} * \dfrac{\sqrt{s(x + h)} + \sqrt{s(x)}}{\sqrt{s(x + h)} + \sqrt{s(x)}} = \dfrac{s(x + h) - s(x)}{h\left(\sqrt{s(x + h)} + \sqrt{s(x)}\right)}.\)
Now you will be able to find a limit to that mess as h approaches 0
if h cancels out of the denominator and s(x) is not zero.
