# Evaluating a limit as x tends to $\infty$

1. May 13, 2015

### Raghav Gupta

1. The problem statement, all variables and given/known data

$$\lim_{x\to\infty} (\sqrt{a^2x^2 +bx + x} -ax) =$$
b/2a
b/a
0
2b/a
2. Relevant equations
Lim x tends to infinity 1/x = 0

3. The attempt at a solution
Taking a2x2 common from square root, we get
$$\lim_{x\to\infty} (ax \sqrt{1+\frac{b}{a^2x}+ \frac{1}{a^2x}} -ax) =$$
Putting x as infinity we get,
ax - ax = 0
But Wolfram is showing something else

2. May 13, 2015

### ehild

Wolfram is right. You multiply something that tends to zero with something that goes to infinity. You can not conclude that the limit is zero.

3. May 13, 2015

### ehild

Use the approximation $\sqrt{1+\delta} = 1+0.5 \delta$ valid if δ<<1.

4. May 13, 2015

### Staff: Mentor

You took the limit too soon. Instead, notice that both terms have a factor of ax, so factor that expression out so that you're taking the limit of ax * <other stuff>. What do you get now for your limit?

5. May 13, 2015

### SammyS

Staff Emeritus
What is the result if a is not a positive number ?

6. May 13, 2015

### Ray Vickson

None of the possible answers listed is correct. What do you get if a > 0? What do you get if a < 0?

7. May 13, 2015

### Raghav Gupta

I get

$$\lim_{x\to\infty} (ax( \sqrt{1+\frac{b}{a^2x}+ \frac{1}{a^2x}} -1))$$
Now what should I do to solve it?
Can some formula, L Hopital rule etc. be applied here?
Don't know for both the cases.
It is amazing that most of the questions which I post are wrong or options are wrong. ( Test makers should verify the questions as it wastes time in exam hour.)
Okay leaving the options,
Can we say for this problem that the limit not exists?

8. May 13, 2015

### Staff: Mentor

$\sqrt{a^2x^2} = |ax|$
Since the limit is as $x \to \infty$, we can assume that x > 0, but since no information is given about a, the best we can do is to write $\sqrt{a^2x^2} = |a|x$.

Regarding the limit above, use the fact that $\lim AB = \lim A \cdot \lim B$, provided that both limits exist.

9. May 13, 2015

### Raghav Gupta

So lim B = 0
Lim A = ∞
LimA.LimB = ∞ * 0 which is indeterminate.

10. May 13, 2015

### SammyS

Staff Emeritus
I suspect the problem should have been $\displaystyle\ \lim_{x\to\infty} (\sqrt{a^2x^2 +bx + c\ } -ax) \$ or similar.

Multiply & divide by the complex conjugate ( ← Added in quick Edit. Mark beat me to it!)

11. May 13, 2015

### Staff: Mentor

You're right. A trick that often works in cases like this is to multiply by the conjugate over itself. (IOW, multiply by 1 in a certain form).

Going back to the original problem, $\lim_{x\to\infty} (\sqrt{a^2x^2 +bx + x} -ax)$, try multiplying by $\frac{\sqrt{a^2x^2 +bx + x} + ax}{\sqrt{a^2x^2 +bx + x} + ax}$

BTW, are you sure about the quantity in the radical? It seems odd to me to have bx + x in there. I'm wondering if the last two terms should be bx + c. Just checking.

Edit: Sammy beat me to it.

12. May 13, 2015

### Ray Vickson

No, you cannot say that. We have
$$\sqrt{a^2 x^2 + bx + x} = |a x| \left( 1 + \frac{b+1}{a^2} \frac{1}{x} \right)^{1/2} - ax \doteq |a|x \left(1 + \frac{b+1}{2a^2} \frac{1}{x} \right) -ax\\ = |a|x - ax + |a| \frac{b+1}{2a^2}.$$
for large $x > 0$. Again, I ask: what do you get if $a > 0$? What do you get if $a < 0$?

Note added in edit: it looks like one of your choices is correct IF a > 0 AND you mean bx + c inside the square root, instead of the bx + x that you wrote.

Last edited: May 13, 2015
13. May 13, 2015

### Raghav Gupta

In the radical there is bx + x term as I have typed earlier. It is sure.
When I do the multiplying and dividing by conjugate and do some solving I get the the answer (b+1)/2a
for a>0. For a<0 it is undefined.

For bx + c term I am getting answer b/2a for a>0
Why wolfram is showing something else?

14. May 13, 2015

### SammyS

Staff Emeritus
It is almost surely a typo -- maybe not your typo, but somebody's typo. (I looked at my keyboard, and right beside the "x" key is the "c" key.)

Tell Wolfram that a>0, (place ", a>0 " immediately after the rest of your input to Wolfram.)

15. May 13, 2015

### Raghav Gupta

Getting not the result, see
wolfram1
And
wolfram2

16. May 13, 2015

### SammyS

Staff Emeritus
The following worked for me: Wolfman .

, a>0

You need the comma .

17. May 13, 2015

### Raghav Gupta

Got it, thanks to all of you- ehild, Sammy,Ray and @Mark44 ( I don't know if it was a coincidence or you are really a fan of Mark 44 ?)

Last edited: May 13, 2015
18. May 13, 2015

### Staff: Mentor

Purely a coincidence - I had never heard of Mark 44 before.

19. May 13, 2015

### BvU

Neither have I
Mark 44 seems a fairly modern contraption. Mark44 is probably in existence since 1944 !

20. May 13, 2015

### Staff: Mentor

Correctamundo!