How to find $\lim_{x\to -\infty} (e^x - 7) (x-a)^2$?

Question

How can I calculate the following limit:

$\lim_{x\to -\infty} (x-a)^2(e^x - 7),\;$ where $a$ is some constant?

Answer 1

So you know that $e^x-7\to-7$ and $x^2 -a\to \infty$. Let's use that to make a formal argument that the limit of their product is $-\infty$. By formal argument, I mean following the definition of "goes to $-\infty$ as $x\to -\infty$".

The definition is that given any real number $M$, there is a point $X$ on the number line such that for any $x<X$ we have $(x^2-a)(e^x-7)< M$.

So let's take an arbitrary real number $M$. By $e^x-7\to-7$, there is some $X_1$ such that for any $x<X_1$, we have $e^x-7<-1$. By $x^2-a\to\infty$, there is an $X_2$ such that for any $x<X_2$ we have $x^2-a > |M|$. Finally, we need to also use the number $X_3$ which is defined as $$ X_3 = \cases{\sqrt a & if $a\geq 0$\\ 0 & otherwise} $$

Now set $X = \min(X_1, X_2, X_3)$. Then any time $x<X$, we have $x<X_1$ and $x<X_2$ and $x<X_3$. By the definitions of the $X_i$ above, these three inequalities imply that $$ (x^2-a)(e^x-7) = -|x^2-a|\cdot |e^x - 7|\\ <-|M|\cdot |-1|\\ \leq M $$ and we're done. (The first equality is because we know the signs of $x^2-a$ and $e^x-7$ thanks to $x<X_1$ and $x<X_3$. The first inequality is using $x<X_1$ and $x<X_2$ and the established inequalities this implies for the two factors.)

Answer 2

HINT

By $y=-x \to \infty$

$$\lim_{x\to -\infty} (e^x - 7) \cdot (x-a)^2=\lim_{y\to \infty} \left(\frac1{e^{y}} - 7\right) \cdot (-y-a)^2=\lim_{y\to \infty} \left(\frac1{e^{y}} - 7\right) \cdot (y+a)^2$$

which is not an indeterminate form.