The Question above states determine the limits, for problem a) I evaluated for when lim x->2 instead of 4. This was marked wrong im simply asking for a clarification on what the question is asking me to do.
Determining Limits
1
$\begingroup$
calculus
limits
3 Answers
2
The "[2, 3, 6]" certainly has nothing to do with what each question is asking. The first clearly says to find the limit as x goes to 4, not 2. The second asks for the limit of $\frac{5x^3+ 40x^2- 45x}{x+ 9}$ as x goes to -9. The first thing you should try is setting x= -9 in this (it worked so nicely for the first problem, didn't it!). Unfortunately that makes the denominator 0 but, fortunately, that also makes the numerator 0 so this limit still might exist. The fact that x= -9 makes the numerator 0 means that x+ 9 is a factor. Divide $5x^3+ 40x^2- 45x$ by $x+ 9$ to find that other factor. Then cancel the "x+ 9" factors.
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[2, 3, 6] might be the number of points given for each correct solution. In any case, it doesn't mean that you should calculate $\lim_{x\to 2}$ instead of $\lim_{x\to 4}$.
So what you should do for a) is
$$\lim_{x\to 4}\sqrt{4x+\sqrt x}=\sqrt{4\cdot 4+\sqrt 4}=\sqrt{18}=3\sqrt 2$$
The first answer is 2 points because you don't have any indeterminate form, just plug in $x=4$. The other answers value more points because there is more work involved to handle the indeterminate forms.
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It is $not$ the answer to any of the questions, but yes it most likely is the points awarded to each question, with $6$ being the hardest question. You evaluated the limit incorrectly, that's why you lost points. If you got $\sqrt{18}$ for the first limit, you would've got $4$ marks