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

[Raghav Gupta]

Homework Statement

$$ \lim_{x\to\infty} (\sqrt{a^2x^2 +bx + x} -ax) = $$
b/2a
b/a
0
2b/a

Homework Equations

Lim x tends to infinity 1/x = 0

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

 

[ehild]

Homework Statement

$$ \lim_{x\to\infty} (\sqrt{a^2x^2 +bx + x} -ax) = $$
b/2a
b/a
0
2b/a

Homework Equations

Lim x tends to infinity 1/x = 0

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

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.

 

[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.

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

 

[Mark44]

Homework Statement

$$ \lim_{x\to\infty} (\sqrt{a^2x^2 +bx + x} -ax) = $$
b/2a
b/a
0
2b/a

Homework Equations

Lim x tends to infinity 1/x = 0

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

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?

But Wolfram is showing something else

 

[SammyS]

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

 

[Ray Vickson]

Homework Statement

$$ \lim_{x\to\infty} (\sqrt{a^2x^2 +bx + x} -ax) = $$
b/2a
b/a
0
2b/a

Homework Equations

Lim x tends to infinity 1/x = 0

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

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

 

[Raghav Gupta]

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?

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?

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

Don't know for both the cases.

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

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?

 

[Mark44]

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?

$\sqrt{a^2x^2} = |ax|$
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.

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?

 

[Raghav Gupta]

$\sqrt{a^2x^2} = |ax|$
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.

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

 

[SammyS]

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!)

 

[Mark44]

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.

 

[Ray Vickson]

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?

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\\<br /> = |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.

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

 

[Raghav Gupta]

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!)

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.

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\\<br /> = |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 + b that you wrote.

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?

 

[SammyS]

In the radical there is bx + x term as I have typed earlier. It is sure.

Why wolfram is showing something else?

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.)

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?

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

 

[Raghav Gupta]

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

Getting not the result, see
wolfram1
And
wolfram2

 

[SammyS]

Getting not the result, see
wolfram1
And
wolfram2

The following worked for me: Wolfman .

, a>0

You need the comma .

 

[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 ?)

 

[Mark44]

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 ?)

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

 

[BvU]

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

 

[Mark44]

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

Correctamundo!

 

[SammyS]

Correctamundo!

I thought maybe Mark44 was 22 times Mark Twain.