# Thread: Limit problem (what's going wrong?)

1. ## Limit problem (what's going wrong?)

${\lim _{x \to - \infty }}{\frac{4x}{\sqrt{x^{2}+4x}+x}}}$

This gives $- \infty$, which is correct.

Dividing both the numerator and the denominator by $x$, we have

${\lim _{x \to - \infty }}{\frac{4}{\sqrt{1+\frac{4}{x}}+1}}}$

now this gives me 2.

What's wrong?

P.S. I tested the values on a calculator.

2. ## Re: Limit problem (what's going wrong?)

Originally Posted by HeilKing
${\lim _{x \to - \infty }}{\frac{4x}{\sqrt{x^{2}+4x}+x}}}$

This gives $- \infty$, which is correct.

Dividing both the numerator and the denominator by $x$, we have

${\lim _{x \to - \infty }}{\frac{4}{\sqrt{1+\frac{4}{x}}+1}}}$

now this gives me 2.

What's wrong?

P.S. I tested the values on a calculator.
Do not double post.

You can be banded for that.

3. ## Re: Limit problem (what's going wrong?)

Originally Posted by HeilKing
${\lim _{x \to - \infty }}{\frac{4x}{\sqrt{x^{2}+4x}+x}}}$

This gives $- \infty$, which is correct.

Dividing both the numerator and the denominator by $x$, we have

${\lim _{x \to - \infty }}{\frac{4}{\sqrt{1+\frac{4}{x}}+1}}}$

now this gives me 2.

What's wrong?

P.S. I tested the values on a calculator.

first it can be written as

$$\mathop {\lim }\limits_{x \to - \infty } \frac{{4x}}{{\sqrt {{x^2} + 4x} + x}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{4x}}{{\sqrt {{x^2} + 4x} + x}}*(\sqrt {{x^2} + 4x} - x)/(\sqrt {{x^2} + 4x} - x) = \mathop {\lim }\limits_{x \to - \infty } \frac{{4x}}{{4x}}*(\sqrt {{x^2} + 4x} - x) = \mathop {\lim }\limits_{x \to - \infty } \sqrt {{x^2} + 4x} - x$$

now you cannot take x out since it is a negative number

this question can be done in the following way

since x is negative , convert it to a positive number by substituting y= -x or x= -y

therefore functon becomes $$\mathop {\lim }\limits_{x \to - \infty } (\sqrt {{x^2} + 4x} - x) = \mathop {\lim }\limits_{y \to \infty } \sqrt {{y^2} - 4y} + y = \mathop {\lim }\limits_{y \to \infty } y(\sqrt {1 - 4/y} + 1) = \mathop {\lim }\limits_{y \to \infty } y(1 + 1) = \mathop {\lim }\limits_{y \to \infty } 2y = \infty$$

thus answer is infinity not minus infinity......