Is there a way to tell if a limit is positive infinity or negative infinity without graphing?

Question

If I have such limit $\displaystyle\lim_{x \to -\infty} \frac{2x^2-4x}{x+1}$ to calculate.

How can I know if the result if $-\infty$ or $\infty$ if I don't have a way to graph this function and I don't know how this graph looks like?

Because the direct substitution will be like this:

• $\dfrac{2(-\infty)^2-4(-\infty)}{-\infty+1}$

I am always confused when calculating the limit when it is approaching $-\infty$ because it is not as easy as ones that approach $\infty$

Answer 1

Assuming the numerator and denominator are polynomials, you can simply ignore all but the highest power term in numerator and denominator, as the highest power term in each will dominate the others as $x$ moves far away from $0$. In your case, for instance, the function behaves like $\frac{2x^2}{x}=2x$ when $x\to\pm \infty$.

One can justify this by, for instance, dividing numerator and denominator by the highest power term of the denominator (including the coefficient if there is one), and working with the result. In your case $$ \frac{2x^2/x-4x/x}{x/x+1/x}=\frac{2x-4}{1+\frac1x} $$ It should now be easier to see what happens in the limit.

Answer 2

HINT: $$ \frac{2x^2-4x}{x+1}=\frac{x(2x-4)}{x(1+(1/x))}=\frac{2x-4}{1+(1/x)}. $$

Answer 3

You can divide everything by $x$ to make your expression simpler.

This means that $\frac{2x^2-4x}{x+1} \frac{x}{x}$ this is multiplying by one and will not change your limit.

Now $\frac{\frac{2x^2-4x}{x}}{\frac{x+1}{x}}$ And the denominator is clearly equal to one when $x$ approaches infinity and negative one when $x$ approaches negative infinity. Your numerator simplified to

$2x-4$

Now it’s easy to see the value of your limit.

Tip: In general, plug in a large number (e.g. $10000$), that will give you the limit on tests

Answer 4

You have $\displaystyle\lim_{x \to -\infty} \frac{2x^2-4x}{x+1}$.

The denominator $2x^2 - 4x$ is domined by $2x^2$ as $x$ becomes larger($\infty$) or smaller($-\infty$), and since the pow of $2x^2$ is even, therefore $2x^2 - 4x \to +\infty$ as $x \to -\infty$

The numerator tends to negative value as $x \to - \infty$.

Therefore the "mixture" of both is an negative sign, therefore the function will have negative sign as $x\to -\infty$.