- #1
atomicpedals
- 202
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The other day in a fit of boredom I decided to dust off my old math books (high school and undergrad) and see if I can still do basic calculus. These days if I need to solve anything I ask a computer to do it, the hazards of getting a job in industry I suppose.
All that said, I have been tripped up by finding limits by conjugates.
1. Homework Statement
Rationalize the following expression by conjugates: $$ \lim_{x \rightarrow -1} \frac { (x+1) } { \sqrt{x+5} - 2}$$
Not applicable.
So the easy way to solve this is to simply plot the function, there you clearly see that the limit is 4. However, that's too easy as this can clearly be done by hand (otherwise the exercise would have said "graph it"). The basic algebra is where I go horribly astray.
$$ \frac { (x+1) } { \sqrt{x+5} - 2} = \frac { (x+1) } { \sqrt{x+5} - 2} \frac { \sqrt{x+5} + 2 } {\sqrt{x+5} + 2} $$
Having cheated and graphed the function I can tell by inspection that the denominator should go to ## (x + 1) ## to cancel out the ## (x + 1) ## in the numerator and yield the limit of 4. And as I'm writing this I think I've answered my own question (huzzah!); it's a simple application of FOIL. $$ ( \sqrt {x + 5} - 2 ) ( \sqrt {x + 5} + 2 ) = x + 5 - 4 = (x + 1) $$ And so $$ \lim_{x \rightarrow -1} \frac { (x+1) } { \sqrt{x+5} - 2} = \lim_{x \rightarrow -1} \sqrt{x+5} + 2 = \sqrt{4} + 2 = 2 + 2 = 4 $$ Have I gone about this the right way or did I just get lucky?
All that said, I have been tripped up by finding limits by conjugates.
1. Homework Statement
Rationalize the following expression by conjugates: $$ \lim_{x \rightarrow -1} \frac { (x+1) } { \sqrt{x+5} - 2}$$
Homework Equations
Not applicable.
The Attempt at a Solution
So the easy way to solve this is to simply plot the function, there you clearly see that the limit is 4. However, that's too easy as this can clearly be done by hand (otherwise the exercise would have said "graph it"). The basic algebra is where I go horribly astray.
$$ \frac { (x+1) } { \sqrt{x+5} - 2} = \frac { (x+1) } { \sqrt{x+5} - 2} \frac { \sqrt{x+5} + 2 } {\sqrt{x+5} + 2} $$
Having cheated and graphed the function I can tell by inspection that the denominator should go to ## (x + 1) ## to cancel out the ## (x + 1) ## in the numerator and yield the limit of 4. And as I'm writing this I think I've answered my own question (huzzah!); it's a simple application of FOIL. $$ ( \sqrt {x + 5} - 2 ) ( \sqrt {x + 5} + 2 ) = x + 5 - 4 = (x + 1) $$ And so $$ \lim_{x \rightarrow -1} \frac { (x+1) } { \sqrt{x+5} - 2} = \lim_{x \rightarrow -1} \sqrt{x+5} + 2 = \sqrt{4} + 2 = 2 + 2 = 4 $$ Have I gone about this the right way or did I just get lucky?