Limit

\(\lim\limits_{x\to 2} x^3 - 3x^2 + 5x - 2 = 2^3 - 3\cdot 2^2 + 5\cdot 2 - 2 = 4\)

\(\lim\limits_{x\to 2} \frac{2x}{3x^3+1} = \frac{2\cdot 2}{3\cdot 2^3 + 1} = \frac{4}{25}\)

\(\lim\limits_{x\to 2} \sqrt{x^2 + 5} = \sqrt{2^2 + 5} = 3\)

\(\lim\limits_{x\to 2} 3^{x+1} = 3^{2 + 1} = 3^3 =27\)

\(\lim\limits_{x\to 2} log_2 x^5 = log_2 2^5 = 5\)

\(\lim\limits_{x\to 3} \frac{2x^2 - 5x - 3}{x-3}\).

If we directly plug \(3\) into the function, we get denominator \(0\). Hence we have to factorize the numerator and then cancel some common factor.

\(\lim\limits_{x\to 3} \frac{2x^2 - 5x - 3}{x-3}\) \(= \lim\limits_{x\to 3} \frac{(2x + 1)(x-3)}{x-3}\) \(= \lim\limits_{x\to 3} 2x + 1 = 2\cdot 3 + 1 = 7\).

\(\lim\limits_{x\to 9} \frac{\sqrt{x} - 3}{x-9}\).

If we directly plug \(3\) into the function, we get denominator \(0\). Also, neither the numerator nor the denominator can be factorzied. However, We note the numerator is a square root function. In this case, we multiply both numerator and denominator by the conjugate of the numerator (because the numerator contains square root symbol. Similary, if the denominator contains square root symbol, then we multiply by the conjugate of the denominator). The conjuate of \(\sqrt{x} - 3\) is \(\sqrt{x} + 3\), so we have

\(\lim\limits_{x\to 9} \frac{\sqrt{x} - 3}{x-9}\)

\(= \lim\limits_{x\to 9} \frac{(\sqrt{x} - 3)(\sqrt{x} + 3)}{(x-9)(\sqrt{x} + 3)}\) \(= \lim\limits_{x\to 9} \frac{(\sqrt{x})^2 - 3^2}{(x-9)(\sqrt{x} + 3)}\) \(= \lim\limits_{x\to 9} \frac{x - 9}{(x-9)(\sqrt{x} + 3)}\)

\(= \lim\limits_{x\to 9} \frac{1}{\sqrt{x} + 3}\) \(= \frac{1}{6}\).

• Description

  Suppose \(f(x) = \frac{g(x)}{h(x)}\) where

  • \(g(x) = a_0 + a_1 x + a_2 x^2 + \dots + a_p x^p\) is a polynomial of order \(p\)
  • \(h(x) = b_0 + b_1 x + b_2 x^2 + \dots + b_q x^q\) is a polynomial of order \(q\)

  To find \(\lim\limits_{x\to\infty} f(x)\) or \(\lim\limits_{x\to -\infty} f(x)\), we need

  • Divide both numerator and denominator by \(x^{max\{p, q\}}\) and simplify the quotients, and
  • Evaluate the limit using the simplified form.

• Example

  \(\lim\limits_{x\to \infty} \frac{2x^2 -3x + 1}{x^2 -5}\)

  \(= \lim\limits_{x\to \infty} \frac{2x^2 -3x + 1}{x^2 -5}\) \(= \lim\limits_{x\to \infty} \frac{\frac{2x^2 -3x + 1}{x^2}}{\frac{x^2 -5}{x^2}}\) \(= \lim\limits_{x\to \infty} \frac{2 - \frac{3}{x} + \frac{1}{x^2}} {1 - \frac{5}{x^2}}\)

  \(= \frac{ \lim\limits_{x\to \infty} 2 - \lim\limits_{x\to \infty} \frac{3}{x} + \lim\limits_{x\to \infty} \frac{1}{x^2}} {\lim\limits_{x\to \infty} 1 - \lim\limits_{x\to \infty} \frac{5}{x^2}}\)

  \(= \frac{2-0+0}{1-0}\) \(= 2\)

• Remarks : Shortcuts
  • If \(p=q\), then \(\lim\limits_{x\to\infty} f(x) = \lim\limits_{x\to-\infty} f(x) = \frac{a_p}{b_q}\).

  • If \(p < q\), then \(\lim\limits_{x\to\infty} f(x) = \lim\limits_{x\to-\infty} f(x) = 0\).

    \(\lim\limits_{x\to\infty} \frac{x^2+1}{x^3-2} = 0\)

  • If \(p > q\), then

    • \(\lim\limits_{x\to\infty} f(x) = \infty\).

      \(\lim\limits_{x\to\infty} \frac{x^3-2}{x^2+1} = \infty\)

    • \(\lim\limits_{x\to -\infty} f(x) =\)

      • \(\infty\) if \(p-q\) is even
      • \(-\infty\) if \(p-q\) is odd.

      \(\lim\limits_{x\to -\infty} \frac{x^3-2}{x^2+1} = -\infty\)

      \(\lim\limits_{x\to -\infty} \frac{x^4-2}{x^2+1} = \infty\)

• Description

  Suppose

  • \(g(x) = a_0 + a_1 x + a_2 x^2 + \dots + a_p x^p\) is a polynomial of order \(p\)
  • \(h(x) = b_0 + b_1 x + b_2 x^2 + \dots + b_q x^q\) is a polynomial of order \(q\)

  Then

  • If \(f(x) = \frac {\sqrt{g(x)}} {h(x)}\), then divide both numerator and denominator by \(max\{p/2, q\}\).
  • If \(f(x) = \frac {g(x)} {\sqrt{h(x)}}\), then divide both numerator and denominator by \(max\{p, q/2\}\).
  • If \(f(x) = \frac {\sqrt{g(x)}} {\sqrt{h(x)}}\), then divide both numerator and denominator by \(max\{p/2, q/2\}\).

• Example

  \(\lim\limits_{x\to \infty} \frac{\sqrt{2x^2 + 1}}{x - 5}\)

  Now we can see

  • \(f(x) = \frac {\sqrt{g(x)}} {h(x)}\)
  • \(p = 2\)
  • \(q = 1\)
  • \(max\{p/2, q\} = max\{ 2/2, 1 \} = 1\)

  so we divide both top and bottom by \(x^1 = x\) :

  \(\lim\limits_{x\to \infty} \frac{\sqrt{2x^2 + 1}}{x - 5}\)

  \(= \lim\limits_{x\to \infty} \frac{\frac{\sqrt{2x^2 + 1}}{x}}{\frac{x -5}{x}}\) \(= \lim\limits_{x\to \infty} \frac {\frac {\sqrt{2x^2 + 1}} {\sqrt{x^2}} } {\frac {x-5} {x} }\) \(= \lim\limits_{x\to \infty} \frac {\sqrt{ \frac {2x^2 + 1} {x^2} } } {1 - \frac {5} {x} }\)

  \(= \lim\limits_{x\to \infty} \frac {\sqrt{ 2 + \frac{1}{x^2} } } {1 - \frac {5} {x} }\) \(= \frac {\lim\limits_{x\to \infty} \sqrt{ 2 + \frac{1}{x^2} } } {\lim\limits_{x\to \infty} 1 - \frac {5} {x} }\) \(= \frac{\sqrt{2+0}}{1-0}\)

  \(= \sqrt{2}\)

  Here we use \(x = \sqrt{x^2}\), which is only true for \(x\geq 0\). For \(x < 0\), \(x = -\sqrt{x^2}\) holds. Since here we are evaluating the limit at \(+\infty\), we consider only positive \(x\). If we evaluate the limit at \(-\infty\), then it should be

  \(\lim\limits_{x\to -\infty} \frac{\sqrt{2x^2 + 1}}{x - 5}\)

  \(= \lim\limits_{x\to -\infty} \frac{\frac{\sqrt{2x^2 + 1}}{x}}{\frac{x -5}{x}}\) \(= \lim\limits_{x\to -\infty} \frac {\frac {\sqrt{2x^2 + 1}} {-\sqrt{x^2}} } {\frac {x-5} {x} }\) \(= \lim\limits_{x\to -\infty} - \frac {\sqrt{ \frac {2x^2 + 1} {x^2} } } {1 - \frac {5} {x} }\)

  \(= \lim\limits_{x\to -\infty} -\frac {\sqrt{ 2 + \frac{1}{x^2} } } {1 - \frac {5} {x} }\) \(= -\frac {\lim\limits_{x\to -\infty} \sqrt{ 2 + \frac{1}{x^2} } } {\lim\limits_{x\to -\infty} 1 - \frac {5} {x} }\) \(= -\frac{\sqrt{2+0}}{1-0}\)

  \(= -\sqrt{2}\)