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For which $a,b\in\mathbb{R}$, is $f(x)$ continuous?

$f:(-1,+\infty)$, $f(x)=\lim\limits_{n\to\infty} \frac{x^2+bx}{a+x^n}$

pre calculus, can't figure out how to analyze continuity at $x=1$

2 Answers 2

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First evaluate the limit:

$$f(x)= \begin{cases} \frac{x^2 + bx}{a}, & x\in(-1,1) \\ \frac{1+b}{1+a}, & x=1 \\ 0, & x>1 \end{cases}$$

To enforce continuity at $x=1$, we must have

$$\frac{1+b}{a} = \frac{1+b}{1+a}=0.$$

This is true for $b=-1$ and any $a\not\in \{-1,0\}.$

  • 0
    I guess my confusion comes from the fact that I can't seem to understand how can we calculate $\lim\limits_{n\to\infty, x\to1^{-}} x^n$ – 2017-01-22
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    The left-hand limit is zero, since for any $x\in(0,1),$ $x^n\to 0$ as $n\to\infty$. Similarly, the right-hand limit is $\infty$, since for any $x>1,$ $x^n\to \infty$. – 2017-01-22
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For any $x>1$, we have that $\lim\limits_{n\to\infty}x^n=+\infty$, and thus $\displaystyle f(x)=\lim_{n\to\infty}\frac{x^2+bx}{a+x^n}=0$. Therefore for continuity at $x=1$, $f(1)$ has to be equal to zero too. Since $\displaystyle f(1)=\frac{b+1}{a+1}$, setting up the equation $\displaystyle \frac{b+1}{a+1}=0$, we'll find the desired value of $b=-1$.

For any $-1