Solving Limits: Finding a & b When x Approaches 0

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In summary, to deduce the values of a and b from the given limit as x approaches 0, the numerator and denominator should be multiplied by the conjugate. This will result in (a+bx-5)/(x√(a+bx)+√5)). Since the denominator cannot be equal to zero, a must equal 5.
  • #1
cripple
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Homework Statement



I'm doing an assignment, and I've hit a wall with this limits question:

6. If the limit as x approaches 0 of [URL]http://www2.wolframalpha.com/Calculate/MSP/MSP202919ggbg4de783c0e7000024ba5hah2cdg59e6?MSPStoreType=image/gif&s=14&w=106&h=41[/URL] equals [itex]\sqrt{5}[/itex], deduce the values of a and b.


The Attempt at a Solution



So far, I've multiplied the numerator and denominator by the conjugate, and now I have [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP231719ggb7fd1fig87b800001ie7f55ge9hc9190?MSPStoreType=image/gif&s=9&w=106&h=42. I've tried squaring the bottom and the top and expanding them out, but I'm just spinning my wheels at this point, I don't really know how to proceed. I would appreciate any hints you could give me, thanks in advance.
 
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  • #2


cripple said:

Homework Statement



I'm doing an assignment, and I've hit a wall with this limits question:

6. If the limit as x approaches 0 of [URL]http://www2.wolframalpha.com/Calculate/MSP/MSP202919ggbg4de783c0e7000024ba5hah2cdg59e6?MSPStoreType=image/gif&s=14&w=106&h=41[/URL] equals [itex]\sqrt{5}[/itex], deduce the values of a and b.

The Attempt at a Solution



So far, I've multiplied the numerator and denominator by the conjugate, and now I have [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP231719ggb7fd1fig87b800001ie7f55ge9hc9190?MSPStoreType=image/gif&s=9&w=106&h=42. I've tried squaring the bottom and the top and expanding them out, but I'm just spinning my wheels at this point, I don't really know how to proceed. I would appreciate any hints you could give me, thanks in advance.

Are the fractions in the images correct? Because if I multiply
[itex](\sqrt{a+bx}-\sqrt{5})(\sqrt{a+bx}+\sqrt{5})[/itex]
I get
a + bx - 5,
and that's not what's in the numerator of the 2nd image.
 
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  • #3


Sorry, I left out this part -[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP157319ggbcc94ih84dd300005ia505e2i20325d5?MSPStoreType=image/gif&s=24&w=126&h=46 - that's what I got after multiplying by the conjugate, then I tried to cancel the x to get rid of the division by zero at the limit
 
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  • #4


cripple said:
[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP231719ggb7fd1fig87b800001ie7f55ge9hc9190?MSPStoreType=image/gif&s=9&w=106&h=42[/QUOTE]

Are you from University of Auckland? My friend from Auckland had the exact same problem. Anyways...

You multiplied incorrectly. After multiplying by the conjugate, you get:

(a+bx-5)/(x√(a+bx)+√5))

The only way this will approach a limit is if you manage to cancel out the x on the bottom. The only way to do that is to get the fraction to have the form x/x.
 
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  • #5


You mean that you tried to cancel the x going from this:
[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP157319ggbcc94ih84dd300005ia505e2i20325d5?MSPStoreType=image/gif&s=24&w=126&h=46
to this?
[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP231719ggb7fd1fig87b800001ie7f55ge9hc9190?MSPStoreType=image/gif&s=9&w=106&h=42
If so, then you can't do that.
 
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  • #6


Harrisonized said:
Are you from University of Auckland? My friend from Auckland had the exact same problem. Anyways...

You multiplied incorrectly. After multiplying by the conjugate, you get:

(a+bx-5)/(x√(a+bx)+√5))

The only way this will approach a limit is if you manage to cancel out the x on the bottom. The only way to do that is to get the fraction to have the form x/x.
I did this step, I should have included it in the OP. anyway yes I am from Auckland University, I am terrible at maths but decided to do 108 anyway, which is starting to look like it may have been a mistake:rofl:
 
  • #7


Thanks eumyang, I'll try again.
 
  • #8


Hint (copied directly from your assignment): Multiply the numerator and denominator by the conjugate and remember that the denominator can not be zero.

The bolded part means that a=5.
 
  • #9


Thanks Harrisonized, didn't realize the hint about the denominator meant a=5
 
  • #10


So how do we deduce that a=5 from that? I am at this step: (a+bx-5)/(x√(a+bx)+√5))
but unsure how to move forward from there.
 

1. What is the purpose of solving limits?

The purpose of solving limits is to determine the value that a function or expression approaches as its input variable gets closer and closer to a specific point. This can provide valuable information about the behavior of the function and help in evaluating complex expressions.

2. How do you find the limit of a function?

To find the limit of a function, you can use algebraic manipulation, graphing, or substitution. In some cases, you may need to apply special limit theorems or techniques such as L'Hopital's rule.

3. What does it mean when x approaches 0 in a limit?

When x approaches 0 in a limit, it means that the input variable of the function is getting closer and closer to the value of 0. This does not necessarily mean that the function is actually evaluated at 0, but rather that we are observing the behavior of the function as the input approaches 0.

4. What is the difference between a and b in finding a limit?

In the context of solving limits, a and b represent the coefficients or constants in an expression that has x approaching 0. a represents the coefficient of the highest power of x, while b represents the constant term. Both a and b play a role in determining the limit of the expression.

5. Can you always find the limit of a function?

In some cases, it may not be possible to find the limit of a function. This can happen when the function is undefined at the point we are evaluating, or when the limit does not exist due to a jump or oscillation in the function's behavior. In these cases, the limit is said to be undefined or does not exist.

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