Limit problem (what's going wrong?)

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

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

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

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

now this gives me 2.

What's wrong?

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

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

Quote:

Originally Posted by

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

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

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

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

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.

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

Quote:

Originally Posted by

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

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

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

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

now this gives me 2.

What's wrong?

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

first it can be written as

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

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 $\displaystyle \[\mathop {\lim }\limits_{x \to - \infty } (\sqrt {{x^2} + 4x} - x) = \mathop {\lim }\limits_{y \to \infty } \sqrt {{y^2} - 4y} + y = \mathop {\lim }\limits_{y \to \infty } y(\sqrt {1 - 4/y} + 1) = \mathop {\lim }\limits_{y \to \infty } y(1 + 1) = \mathop {\lim }\limits_{y \to \infty } 2y = \infty \]$

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