pinkphiloyd
New member
- Joined
- Oct 6, 2015
- Messages
- 9
First, here's my backstory. Feel free to skip this paragraph if you'd like. I'm going back to school for a second Bachelor's beginning in January. I took Cal I as an elective the first time around and did well, but that was in 2009. My intention was to review through trig and take Cal I again, but after talking with my advisor, his advice was to review through Cal I and pick up immediately at Cal II, as this would allow me to finish my new pursuit in 2 years as opposed to 3. So, here I am. Trying to relearn Calculus after having done nothing but some basic Algebra since 2009. I've been using Khan academy to review. (I have ordered a couple of well reviewed texts to use as well, but they haven't arrived yet.) I started Algebra and Trig there, worked for a couple of weeks, and then decided it would be more efficient to move to the calculus and review other topics as they came up and I needed them. Maybe that's my mistake, although I actually feel like it's been going pretty well so far.
Okay, with that out of the way, here's my question. I feel like I'm missing something really, really obvious here and I'm gonna kick myself after I ask.
I encountered this problem on Khan academy. I came up with the right answer, but I cheated by determining that answer graphically. After doing so, I clicked the hints so that I could see and review what was actually going on.
Try multiplying by a form of 1 using the conjugate of the radical expression:
. . . . .\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, \infty}\, \left(\, \sqrt{\strut 9x^2\, +\, 4x\,}\, -\, 3x\right)\, \cdot\, \left( \dfrac{\sqrt{\strut 9x^2\, +\, 4x\,}\, +\, 3x}{\sqrt{\strut 9x^2\, +\, 4x\,}\, +\, 3x}\right)\)
Simplify this rather complicated-looking expression:
. . . . .\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, \infty}\, \dfrac{ (9x^2\, +\, 4x)\, -\, 9x^2}{\sqrt{\strut 9x^2\, +\, 4x\,}\, +\, 3x}\, =\, \) \(\displaystyle \,\color{blue}{\lim_{x\, \rightarrow \, \infty}\, \dfrac{4x}{\sqrt{\strut 9x^2\, +\, 4x\,}\, +\, 3x}}\)
Now divide the numerator and the denominator by 1 in the form of \(\displaystyle \,\color{red}{\mathbf{\mathit{\dfrac{x}{x}:}}}\)
. . . . .\(\displaystyle \displaystyle \color{red}{ \mathbf{ \lim_{x\, \rightarrow\, \infty}\, \dfrac{ \dfrac{4\mathit{x}}{\mathit{x}} }{ \dfrac{ \sqrt{\strut 9\mathit{x}^2\, +\, 4\mathit{x}\,} }{\mathit{x}}\, +\, \dfrac{3\mathit{x}}{\mathit{x}} } }}\, \) \(\displaystyle \, =\, \lim_{x\, \rightarrow\, \infty}\, \dfrac{4}{\dfrac{ \sqrt{\strut 9x^2\, +\, 4x\,} }{x}\, +\, 3 }\)
Recall from algebra that \(\displaystyle \, \sqrt{\strut x^2\,}\, =\, x\, \) for positive values of \(\displaystyle \, x.\)
. . . . .\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, \infty}\, {\dfrac{ \sqrt{\strut 9x^2\, +\, 4x\,} }{x}\, +\, 3 }\, =\, \lim_{x\, \rightarrow\, \infty}\, \dfrac{4}{ \sqrt{\strut \dfrac{9x^2}{x^2}\, +\, \dfrac{4x}{x^2}\,}\, +\, 3 }\, =\, \lim_{x\, \rightarrow\, \infty}\, \dfrac{4}{\sqrt{\strut 9\, +\, \dfrac{4}{x}\,}\, +\, 3}\)
As \(\displaystyle \,x\,\) increases without bound, \(\displaystyle \, \dfrac{4}{x}\, \) approaches \(\displaystyle \, 0.\,\) Thus we have:
. . . . .\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, \infty}\, \dfrac{4}{\sqrt{\strut 9\, +\, \dfrac{4}{x}\,}\, +\, 3}\, =\, \dfrac{4}{\sqrt{\strut 9\,+\,0\,}\, +\, 3}\, =\, \dfrac{4}{\sqrt{\strut 9\,}\, +\, 3}\, =\, \dfrac{4}{3\, +\, 3}\, =\, \dfrac{4}{6}\, =\, \dfrac{2}{3}\)
So. I'm good with everything except the bit in red. I can get to the bit in blue, no problem. I understand that the part circled in blue needs further simplification. After we do the bit circled in red (divide by x/x), I can manipulate the problem the rest of the way and get you to the solution. I'm just fuzzy on why we divided by x/x. How did we know that that's what we needed to do? I appreciate the help.
Okay, with that out of the way, here's my question. I feel like I'm missing something really, really obvious here and I'm gonna kick myself after I ask.
I encountered this problem on Khan academy. I came up with the right answer, but I cheated by determining that answer graphically. After doing so, I clicked the hints so that I could see and review what was actually going on.
Try multiplying by a form of 1 using the conjugate of the radical expression:
. . . . .\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, \infty}\, \left(\, \sqrt{\strut 9x^2\, +\, 4x\,}\, -\, 3x\right)\, \cdot\, \left( \dfrac{\sqrt{\strut 9x^2\, +\, 4x\,}\, +\, 3x}{\sqrt{\strut 9x^2\, +\, 4x\,}\, +\, 3x}\right)\)
Simplify this rather complicated-looking expression:
. . . . .\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, \infty}\, \dfrac{ (9x^2\, +\, 4x)\, -\, 9x^2}{\sqrt{\strut 9x^2\, +\, 4x\,}\, +\, 3x}\, =\, \) \(\displaystyle \,\color{blue}{\lim_{x\, \rightarrow \, \infty}\, \dfrac{4x}{\sqrt{\strut 9x^2\, +\, 4x\,}\, +\, 3x}}\)
Now divide the numerator and the denominator by 1 in the form of \(\displaystyle \,\color{red}{\mathbf{\mathit{\dfrac{x}{x}:}}}\)
. . . . .\(\displaystyle \displaystyle \color{red}{ \mathbf{ \lim_{x\, \rightarrow\, \infty}\, \dfrac{ \dfrac{4\mathit{x}}{\mathit{x}} }{ \dfrac{ \sqrt{\strut 9\mathit{x}^2\, +\, 4\mathit{x}\,} }{\mathit{x}}\, +\, \dfrac{3\mathit{x}}{\mathit{x}} } }}\, \) \(\displaystyle \, =\, \lim_{x\, \rightarrow\, \infty}\, \dfrac{4}{\dfrac{ \sqrt{\strut 9x^2\, +\, 4x\,} }{x}\, +\, 3 }\)
Recall from algebra that \(\displaystyle \, \sqrt{\strut x^2\,}\, =\, x\, \) for positive values of \(\displaystyle \, x.\)
. . . . .\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, \infty}\, {\dfrac{ \sqrt{\strut 9x^2\, +\, 4x\,} }{x}\, +\, 3 }\, =\, \lim_{x\, \rightarrow\, \infty}\, \dfrac{4}{ \sqrt{\strut \dfrac{9x^2}{x^2}\, +\, \dfrac{4x}{x^2}\,}\, +\, 3 }\, =\, \lim_{x\, \rightarrow\, \infty}\, \dfrac{4}{\sqrt{\strut 9\, +\, \dfrac{4}{x}\,}\, +\, 3}\)
As \(\displaystyle \,x\,\) increases without bound, \(\displaystyle \, \dfrac{4}{x}\, \) approaches \(\displaystyle \, 0.\,\) Thus we have:
. . . . .\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, \infty}\, \dfrac{4}{\sqrt{\strut 9\, +\, \dfrac{4}{x}\,}\, +\, 3}\, =\, \dfrac{4}{\sqrt{\strut 9\,+\,0\,}\, +\, 3}\, =\, \dfrac{4}{\sqrt{\strut 9\,}\, +\, 3}\, =\, \dfrac{4}{3\, +\, 3}\, =\, \dfrac{4}{6}\, =\, \dfrac{2}{3}\)
So. I'm good with everything except the bit in red. I can get to the bit in blue, no problem. I understand that the part circled in blue needs further simplification. After we do the bit circled in red (divide by x/x), I can manipulate the problem the rest of the way and get you to the solution. I'm just fuzzy on why we divided by x/x. How did we know that that's what we needed to do? I appreciate the help.
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