Calculus please tell me if I am right

Question

$$ f(x) = \begin{cases} x^2 &\text{for } x > 1 \\ x-1 &\text{for } 0 < x \le 1 \\ 0 &\text{for } x \le 0 \end{cases} $$ Then what is the value of $\lim_{x\to 0^-}f(x^2) +\lim_{x\to 1^+} \frac{f(x-1)}{x+2} $ ?

The answer key is $-4/3$ but when I try to solve this, I got $0$.

Am I right? Can anyone tell me please.

Answer 1

Note that

$$\lim_{x\to 0^-}f(x^2)=\lim_{y\to 0^+}f(y)=-1$$

$$\lim_{x\to 1^+} \frac{f(x-1)}{x+2}=\lim_{y\to 0^+} \frac{f(y)}{3+y}=-\frac13$$

then

$$\lim_{x\to 0^-}f(x^2)+\lim_{x\to 1^+} \frac{f(x-1)}{x+2}=-1-\frac13=-\frac43$$

Answer 2

Consider this:

If $-1 < x < 0$ then $0 < x^2 < 1$ and $f(x^2) = x^2-1$ and $\lim_{x\to 0^-} f(x^2)=\lim_{x\to 0^-}x^2 -1 =-1$.

and if $1 < x < 2$ then $0 < x - 1< 2$ and $f(x-1) = x-1 -1 = x-2$ and $\frac {f(x-1)}{x + 2}=\frac {x-2}{x+2}$. So $\lim_{x\to 1^+}\frac {f(x-1)}{x+2} =\lim_{x\to 1^+}\frac {x-2}{x+2} = -\frac 13$.

And so $\lim_{x\to 0^-} f(x^2)+ \lim_{x\to 1^+}\frac {f(x-1)}{x+2} = -1 + (-\frac 13) = -\frac 43$.

.....

Your mistake was assuming $\lim_{x\to 0^-}$ would mean $\lim_{x^2\to 0^-}$ and $\lim_{x \to 1^+} $ would mean $\lim_{x-1 \to 1^+}$. (the second one when you think about it doesn't actually make any sense, and the first one is just wrong.)

In actuality $\lim_{x \to 0^-}$ means $\lim_{x^2 \to 0^+}$ and $\lim_{x\to 1^+}$ means $\lim_{x-1 \to 0^+}$.

....

Another way to explain your error:

Let $g_1(x) = x^2; g_2(x) = x -1; g_3(x) = 0$ and $f(x) = \begin{cases} g_1(x) &\text{for } x > 1 \\ g_2(x) &\text{for } 0 < x \le 1 \\ g_3(x) &\text{for } x \le 0 \end{cases}$

You assumed that if $x < 0$ then $f(x^2) = g_3(x^2)$ and if $x > 1$ then $f(x-1) = g_1(x -1)$.

But $x < 0$ means $x^2 > 0$ and if $-1 < x < 0$ then $0 < x^2 < 1$ so $f(x^2) = g_2(x)$. Likewise if $2 > x > 1$ then $0 < x - 1 < 1$ so $f(x-1) = g_2(x-1)$.