calculusdifferentiallimits
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:
I am always confused when calculating the limit when it is approaching $-\infty$ because it is not as easy as ones that approach $\infty$
Hint: Let $t=\dfrac1x$ , and use the fact that $\displaystyle\lim_{a\to\infty}x^a=0$ when $x\in(0,1)$.
Why there is no limit?
The graphic can help you understand why and suggest you some approach for the proof:
Remark: You have to be careful with tables of values because they can be misleading:
\begin{array}{ c | c c c c } x & \frac{1}{2\pi} & \frac{1}{3\pi} & \frac{1}{4\pi} &\frac{1}{5\pi} \\ \hline \sin\left(\frac{1}{x}\right) & 0 & 0 & 0 & 0 \\ \end{array}
\begin{array}{ c | c c c c } x & \frac{2}{5\pi} & \frac{2}{9\pi} & \frac{2}{13\pi} &\frac{2}{17\pi} \\ \hline \sin\left(\frac{1}{x}\right) & 1 & 1 & 1 & 1 \\ \end{array}
(The tables above are a sketch of the proof - see Theorem 2.4 here.)
Best Answer
HINT: $$ \frac{2x^2-4x}{x+1}=\frac{x(2x-4)}{x(1+(1/x))}=\frac{2x-4}{1+(1/x)}. $$