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# Find the limit or show that it does not exist.$\displaystyle \lim_{x \to -\infty} \left(\sqrt{4x^2 + 3x} + 2x \right)$

## $$\lim _{x \rightarrow-\infty}\left(\sqrt{4 x^{2}+3 x}+2 x\right)=-\frac{3}{4}$$

Limits

Derivatives

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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### Video Transcript

to find this limit. We first want to rationalize this. So we multiply this by the conjugate of this one. That would be the square it of four X squared plus three x minus two X. This all over the same expression square it of four X squared plus three x minus two X. And so from here we have the limit as X approaches negative infinity of we have the square of the square it or four X squared plus three X minus the square of two X. This all over square it of four X squared plus three x -2X. Now simplifying, we have the limit as X approaches negative infinity of four X squared plus three X minus for X squared this all over square it of four X squared plus three x minus two X. I'm simplifying further, we have the limit as X approaches in negative infinity of three x. Over we have square it of four X squared plus three x -2 x. and then and here we factor out the variable with the highest exponent for the denominator. And we have limit as X approaches negative infinity of three X. This all over before the square root we have squared of X squared times four plus three over x -2X. And this gives us limits as X approaches negative infinity of three x over negative X times the square of four plus three X. Now this is negative because our excess approaching negative infinity. So square root of x square This is the same as negative X squared and taking the square it off. So this should be negative X. And then from here We have -2 x. And then we factor out negative X in the denominator. So we have the limit as X approaches negative infinity of three X over you have negative X times square it of four plus three over x minus negative two. Or there will be plus two. And so we can cancel out the X. And we have the limit as X approaches negative infinity of negative three over square at the four plus three X plus two. And here, yeah, evaluate as infinity and we get negative three over Square it of 4-plus 3 over infinity plus to this will give us negative three over the square root of four plus zero plus two, which will give us negative three over two plus two or negative 3/4. And so this is the value of the limits.

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#### Topics

Limits

Derivatives

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

Join Bootcamp