# Evaluating Limits Algebraically

1. Jan 31, 2008

### kwikness

I've never simplified an expression with a subtraction of radicals in the numerator. I've done a bit of reading and I'm guessing the method to use would be conjugation, but I'm not sure how that would let me solve for x. Thanks for any help.

$$\lim_{x\rightarrow 0} \frac{(\sqrt{x+5} - \sqrt{5})}x$$

I'm also having trouble with these two limit estimations. I've worked with $$\Delta$$ before but never in a simplification.

$$\lim_{x\rightarrow 0} \frac{2(x + \Delta x)-2x}{\Delta x}$$

$$\lim_{x\rightarrow 0} \frac{(x + \Delta x^2) - 2(x + \Delta x) + 1 - (x^2 - 2x + 1)}{\Delta x}$$

Last edited: Jan 31, 2008
2. Jan 31, 2008

### rocomath

Multiply both numerator and denominator by ...

$$(\sqrt{x+5}+\sqrt 5)$$

3. Jan 31, 2008

### HallsofIvy

Staff Emeritus
As you suggested and rocophysics confirmed, multiply both numerator and denominator by the conjugate.

Go ahead and do the algebra! This is
$$\lim_{x\rightarrow 0}\frac{2x+ 2\Delta x- 2x}{\Delta x}= \lim_{x\rightarrow 0}\frac{2\Delta x}{\Delta x}[/itex] What is the limit of that? First, you have that typed wrong: it should be $(x+ \Delta x)^2$, not $(x+ \Delta x^2)$! Again, just do the algebra: [tex]\lim_{x\rightarrow 0}\frac{x^2+ 2x\Delta x+ \Delta x^2- 2x- \Delta x+ 1- x^2+ 2x- 1}{\Delta x}$$
$$= \lim_{x\rightarrow 0}\frac{2x\Delta x+ \Delta x^2k}{\Delta x}$$
What is the limit of that?

4. Jan 31, 2008

### kwikness

Thanks very much for your help Ivy.

I can get up to step 4, but how do I go from 4 -> 5? What rule/formula is being used?

Last edited: Jan 31, 2008
5. Jan 31, 2008

### ircdan

multiply by sqrt(5)/sqrt(5)

6. Jan 31, 2008

### kwikness

Wouldn't that leave me with $$\frac{\sqrt{5}}{2}$$? Could you be a bit more specific?

7. Jan 31, 2008

### NateTG

$$2\sqrt{5} \times \sqrt{5}=2 (\sqrt{5})^2=2 (5) = 10$$

8. Jan 31, 2008

### kwikness

Gotchya, I was trying to cancel it out rather than square it. Thanks for the help.

9. Feb 3, 2009

### Liquid7800

To expand on this topic...
Ive not seen many examples of evaluating limits algebraically where there are indexes in the root...for example

$$\lim_{x\rightarrow 64} \frac{\sqrt{x} - 8}{\sqrt[3]{x}-4}$$

....so...are there any good techniques or method to solve this, since you can't use the conjugate?

10. Feb 3, 2009

### JG89

$$\lim_{x\rightarrow 64} \frac{\sqrt{x} - 8}{\sqrt[3]{x}-4} = \lim_{x\rightarrow 64} \frac{(x^{1/6} - 2)(x^{1/3} + 2x^{1/6} + 4)}{(x^{1/6} - 2)(x^{1/6} + 2)} = \lim_{x\rightarrow 64} \frac{x^{1/3} + 2x^{1/6} +4}{x^{1/6} +2}$$.

Now you can evaluate the limit by substituting 64 for x.

11. Feb 4, 2009

### Liquid7800

Thanks for the reply...thats much faster it seems than the way I came up with...even though the principles are the same:

I take the LCM of the exponents
and apply a substitution variable

$$a^6 = 1$$

$$\lim_{a\rightarrow 1} \frac{\sqrt{a^6} - 8}{\sqrt[3]{a^6}-4}$$

Then it kinda becomes a binomial theorem expansion problem:

$$\lim_{a\rightarrow 1} \frac{a^3 - 8}{a^2 - 4}$$

this seems drawn out compared to your way... sorry for the elementary question but how did you decompose the radicals and get the equation

$$\lim_{x\rightarrow 64} \frac{(x^{1/6} - 2)(x^{1/3} + 2x^{1/6} + 4)}{(x^{1/6} - 2)(x^{1/6} + 2)}$$

to simplify to substitute 64 for x?

Again, thanks for the reply!

12. Feb 4, 2009

### JG89

I used a difference of cubes:

$$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$

13. Feb 17, 2009

### Liquid7800

I showed your method for solving that problem to my instructor...she responded by giving me a problem that I am finding rather difficult to implement your method.
Do you think you could help?
Same type of problem, just bigger numbers and no special binomials I could find:

$$\lim_{x\rightarrow 4095} \frac{\sqrt{x+1}- 4\sqrt[3]{x+1}}{\sqrt[4]{x+1}-8}$$

Thanks again.