# Find $\lim_{x\rightarrow 1^{\pm}}\big({\frac {2x+3}{x^2-1}}\big)$

I want to find

$$\lim_{x\rightarrow 1^{\pm}}\left({\frac {2x+3}{x^2-1}}\right)$$

This is what my textbook does:

$$\frac{2x+3}{x^2-1}=\frac{2x+3}{(x-1)(x+1)}=\frac 1 {x-1}\cdot\frac{2x+3}{x+1}$$

Now, for $x\rightarrow1$, we have

$$\frac{2x+3}{x+1}\rightarrow \frac 5 2$$

For $x\rightarrow1^{\pm}$ we have

$$\frac{1}{x-1}\rightarrow\pm\infty$$

This means that

$${\frac {2x+3}{x^2-1}} \rightarrow \pm\infty$$

$${\frac {2x+3}{x^2-1}}\sim \frac{2x}{x^2} = \frac {2}{x}$$

Then I would "replace" (not sure if this is something I can do) $x$ with $1$ (it's the number which the limit approaches to):

$$\frac 2 1 \rightarrow2$$

The limit would be $2$. Replacing $x$ with the number that the limit approaches I think is something I learnt in high school. Any hints on why my reasoning is wrong?

• How does $x$ depend on $n$? – Mark Viola Aug 6 '18 at 14:55
• What is $n$ here? – prog_SAHIL Aug 6 '18 at 14:55
• Sorry @MarkViola, fixed – Cesare Aug 6 '18 at 14:56
• @Cesare Look at the title. Fix that too please. – Mark Viola Aug 6 '18 at 14:57
• Your approach would work if you were looking at the limit as $x\to\infty$. That is where your approach fails here: since $x\to1^{\pm}$, we don't have $\frac{2x+3}{x^2-1}\sim\frac{2x}{x^2}$. – Clayton Aug 6 '18 at 15:03

There is no indeterminant form here.

So you can directly evaluate the limits by putting $x=1^+$ and $x=1^-$

You cannot approximate $\frac{2x+3}{x^2-1}\approx\frac{2}{x}$. It just makes no sense to do so.

The idea $\frac{2x+3}{x^2 - 1} \sim \frac{2x}{x^2}$ works only when the limit is to (positive) infinity. This is because we write $\frac{2x+3}{x^2-1} = \frac{2x+3}{2x} \cdot \frac{x^2}{x^2-1} \cdot \frac{2x}{x^2}$, and when we take the limit as $x$ goes to infinity, we find that the first two fractions have limit $1$, hence the answer is the limit of $\frac{2x}{x^2}$, if the limit were to infinity.

On the other hand, when the limit is as $x \to 1^{+}$, this approach does not work, simply because the two fractions which had limit one earlier, don't have that limit anymore, so we cannot use the product rule to simplify our work.

On the other hand, your "high school" approach helps you in thinking about what the answer could be. In this case, the high school approach substitution on $\frac{2x+3}{x^2 - 1}$ gives $\frac 50$. Therefore, one would guess that the limit does not exist (or it is positive infinite).

The best way to prove this, is to do what the book does : $\frac{2x+3}{x^2-1} = \frac{1}{x-1} \times \frac{2x+3}{x+1}$. In the second expression, substitution works in computation of the limit, and the first has positive infinite as the limit, hence we get the answer.

Suppose you really wanted to use an approach using asymptotics. Then, set $y = \frac 1{x-1}$. Then, as $x \to 1^+$ we have $y \to + \infty$.

We get $\frac{2x+3}{(x-1)(x+1)} = y \left(2 + \frac{1}{x+1}\right) = y \left(2 + \frac y{2y + 1}\right)$.

Now, $\frac{y}{2y+1} \sim \frac 12$, hence $2 + \frac y{2y+1} \sim 2.5$. We still have one $y$ remaining : this contributes to the limit being $+\infty$.

I think you cannot do $\frac{2x+3}{x^2-1} ∼ \frac{2x}{x^2}$. You can do this only when $x$ is big enough so that finite number can be considered as 0 comparing to $x$. But in your case, $x$ approaches to finite number. Thus, the approximation does not work.

Let consider $x=y+1\to 0$ with $y\to 0$

$$\lim_{x\rightarrow 1^{\pm}}{\frac {2x+3}{x^2-1}}=\lim_{y\rightarrow 0} {\frac {y+5}{y(y+3)}}=\lim_{y\rightarrow 0} \frac 1{y} \cdot \lim_{y\rightarrow 0} {\frac {y+5}{y+3}}=\frac53 \lim_{y\rightarrow 0} \frac 1{y}=\pm \infty$$

For $x\rightarrow1^{+}$, we have

$$\frac{2x+3}{x+1}\rightarrow \frac52\quad\text{and}\quad\frac{1}{x-1}\rightarrow+\infty$$

and for $x\rightarrow1^{-}$, we have

$$\frac{2x+3}{x+1}\rightarrow \frac52\quad\text{and}\quad\frac{1}{x-1}\rightarrow-\infty$$

The step $$\frac {2x+3}{x^2-1}\sim \frac{2x}{x^2} = \frac{2}{x}$$ is completely unjustified and totally wrong.

For $x$ near $1$, $2x+5$ is not near $2x$, nor is $x^2-1$ near $x^2$.

You're confusing this with the limit at $\pm\infty$, where those steps are possible and meaningful.