Limits

Find the slope of the secant line to \( k(x) = x^2 \) through points \( (1,1) \) and \( (2,4) \): \[ m_{\text{sec}} = \frac{4-1}{2-1} = 3. \]

Estimate the area under \( f(x) = x^2 + 1 \) over \( [0,3] \) using three rectangles: \[ \text{Area} \approx \sum_{i=1}^{3} \text{Area of Rectangle}_i \]

Estimate \[ \lim_{x \to 3} \frac{x^2 + 5x - 18}{x^2 - 3x}. \]

Define \[ g(x) = \begin{cases} \dfrac{x^2 + 5x - 18}{x^2 - 3x}, & x \neq 3\\ 7, & x=3 \end{cases} \] Estimate \(\lim_{x \to 3} g(x)\).

Answer: As in Example 1, the limit is 4, even though \(g(3) = 7\). This shows that limits depend on nearby values of \(x\), not the function's value at the point itself.

Estimate \[ \lim_{\theta \to 0} \frac{1 - \cos(2\theta)}{\theta}. \]

Observation: The function is undefined at \(\theta = 0\), but by evaluating values close to 0, we can approximate the limit.

Estimate \[ \lim_{t \to 0} \sin(\pi t/2). \]

Observation: Tables of values may help, but they can be misleading. Graphical analysis or algebraic methods provide more reliable estimates.

Estimate \[ \lim_{t \to 0} X(t), \quad X(t) = \begin{cases} -1, & t<0\\ 2, & t \ge 0 \end{cases} \]

Observation: Not all limits exist. Here, the left-hand limit is \(-1\) and the right-hand limit is 2, so the limit does not exist.

Estimate: \[ \lim_{t \to 0^+} H(t) \quad \text{and} \quad \lim_{t \to 0^-} H(t), \quad H(t)=\begin{cases}0,& t<0\\1,& t\ge 0\end{cases} \]

Solution: \[ \lim_{t \to 0^+} H(t) = 1, \quad \lim_{t \to 0^-} H(t) = 0 \]

Estimate: \[ \lim_{t \to 0^+} \sin\left(\frac{1}{t}\right) \quad \text{and} \quad \lim_{t \to 0^-} \sin\left(\frac{1}{t}\right) \]

Observation: The function oscillates infinitely near \(t=0\), so the one-sided limits do not exist.

Let \[ g(x) = \begin{cases} \dfrac{x^2 + 5x -18}{x^2 -3x}, & x\neq 3\\ 7, & x=3 \end{cases} \]

Estimate: \[ \lim_{x \to 3^+} g(x) \quad \text{and} \quad \lim_{x \to 3^-} g(x) \]

Solution: Both one-sided limits exist and equal 4.

Evaluate the following limits: \[ \lim_{x \to 0^+} \frac{1}{x}, \quad \lim_{x \to 0^-} \frac{1}{x}, \quad \lim_{x \to 0} \frac{1}{x} \]

Solution: - As \(x \to 0^+\), \(x>0\) so \(\frac{1}{x} \to +\infty\). - As \(x \to 0^-\), \(x<0\) so \(\frac{1}{x} \to -\infty\). - Since the left and right limits are not equal, \(\lim_{x \to 0} \frac{1}{x}\) does not exist.

Evaluate the following limits: \[ \lim_{x \to 0^+} \frac{6}{x^2}, \quad \lim_{x \to 0^-} \frac{6}{x^2}, \quad \lim_{x \to 0} \frac{6}{x^2} \]

Solution: - As \(x \to 0\), \(x^2>0\) so \(\frac{6}{x^2} \to +\infty\) from both sides. - Therefore, \(\lim_{x \to 0^+} \frac{6}{x^2} = \lim_{x \to 0^-} \frac{6}{x^2} = \lim_{x \to 0} \frac{6}{x^2} = +\infty\)

Evaluate the following limits: \[ \lim_{x \to -2^+} \frac{-4}{x+2}, \quad \lim_{x \to -2^-} \frac{-4}{x+2}, \quad \lim_{x \to -2} \frac{-4}{x+2} \]

Solution: - As \(x \to -2^+\), \(x+2>0\) so \(\frac{-4}{x+2} \to -\infty\). - As \(x \to -2^-\), \(x+2<0\) so \(\frac{-4}{x+2} \to +\infty\). - Left and right limits are not equal, so the limit at \(x=-2\) does not exist.

Evaluate the following limits: \[ \lim_{x \to 4^+} \frac{3}{(4-x)^3}, \quad \lim_{x \to 4^-} \frac{3}{(4-x)^3}, \quad \lim_{x \to 4} \frac{3}{(4-x)^3} \]

Solution: - As \(x \to 4^+\), \(4-x < 0\), so \((4-x)^3 <0\) and \(\frac{3}{(4-x)^3} \to -\infty\). - As \(x \to 4^-\), \(4-x >0\), so \(\frac{3}{(4-x)^3} \to +\infty\). - Since left and right limits differ, \(\lim_{x \to 4} \frac{3}{(4-x)^3}\) does not exist.

Evaluate the following limits: \[ \lim_{x \to 3^+} \frac{2x}{x-3}, \quad \lim_{x \to 3^-} \frac{2x}{x-3}, \quad \lim_{x \to 3} \frac{2x}{x-3} \]

Solution: - As \(x \to 3^+\), \(x-3>0\) so \(\frac{2x}{x-3} \to +\infty\). - As \(x \to 3^-\), \(x-3<0\) so \(\frac{2x}{x-3} \to -\infty\). - Left and right limits differ, so \(\lim_{x \to 3} \frac{2x}{x-3}\) does not exist.

Evaluate the one-sided limit: \[ \lim_{x \to 0^+} \ln x \]

Solution: - As \(x \to 0^+\), \(\ln x \to -\infty\). - So, \(\lim_{x \to 0^+} \ln x = -\infty\)

Evaluate the following limits: \[ \lim_{x \to \pi/2^+} \tan x, \quad \lim_{x \to \pi/2^-} \tan x \]

Solution: - As \(x \to \pi/2^-\), \(\tan x \to +\infty\) - As \(x \to \pi/2^+\), \(\tan x \to -\infty\)

Show that a vertical asymptote exists at \(x=0\) for \(f(x) = 1/x\).

Solution: - One-sided limits: \(\lim_{x\to 0^+} 1/x = +\infty\), \(\lim_{x\to 0^-} 1/x = -\infty\) - Since at least one one-sided limit goes to \(\pm\infty\), \(x=0\) is a vertical asymptote.

Evaluate the following limit: \[ \lim_{x \to 0} \frac{5x+1}{x^2} \]

Solution: - Split: \(\frac{5x+1}{x^2} = \frac{5x}{x^2} + \frac{1}{x^2} = \frac{5}{x} + \frac{1}{x^2}\) - As \(x \to 0^+\), \(\frac{5}{x} + \frac{1}{x^2} \to +\infty\) - As \(x \to 0^-\), \(\frac{5}{x} + \frac{1}{x^2} \to -\infty + \infty = +\infty\) (dominant term \(1/x^2\)) - So, \(\lim_{x \to 0} \frac{5x+1}{x^2} = +\infty\)

Evaluate the following limit: \[ \lim_{x \to -1^+} \frac{2}{x+1}, \quad \lim_{x \to -1^-} \frac{2}{x+1} \]

Solution: - As \(x \to -1^+\), \(x+1>0 \implies \frac{2}{x+1} \to +\infty\) - As \(x \to -1^-\), \(x+1<0 \implies \frac{2}{x+1} \to -\infty\)

\[ \lim_{x \to \infty} (2x^4 - x^2 - 8x), \quad \lim_{t \to -\infty} \left(\frac{1}{3} t^5 + 2 t^3 - t^2 + 8 \right) \]

Solution:

• \(\lim_{x \to \infty} (2x^4 - x^2 - 8x) = \infty\) (dominant \(2x^4\)).

• \(\lim_{t \to -\infty} (\frac{1}{3} t^5 + \dots) = -\infty\) (dominant \(\frac{1}{3}t^5\) negative).

\[ \lim_{x \to \pm \infty} \frac{2x^4 - x^2 + 8x}{-5x^4 + 7} \]

Solution: Factor \(x^4\): \[ \frac{2 - 1/x^2 + 8/x^3}{-5 + 7/x^4} \to -\frac{2}{5} \quad \text{as } x \to \pm \infty \]

\[ \lim_{x \to \infty} \frac{\sqrt{3x^2 + 6}}{5 - 2x}, \quad \lim_{x \to -\infty} \frac{\sqrt{3x^2 + 6}}{5 - 2x} \]

Solution:

• \(x \to \infty: \frac{\sqrt{x^2(3 + 6/x^2)}}{x(-2 + 5/x)} \to -\frac{\sqrt{3}}{2}\)

• \(x \to -\infty: \frac{-x\sqrt{3 + 6/x^2}}{x(-2 + 5/x)} \to \frac{\sqrt{3}}{2}\)

\[ \lim_{x \to \infty} e^{2 - 4x - 8x^2} = 0, \quad \lim_{t \to -\infty} e^{t^4 - 5 t^2 + 1} = \infty, \quad \lim_{z \to 0^+} e^{1/z} = \infty \]

\[ \lim_{x \to \infty} \left(e^{10x} - 4 e^{6x} + \dots \right) = \infty, \quad \lim_{x \to -\infty} \left(e^{10x} - 4 e^{6x} + \dots \right) = -\infty \]

\[ \lim_{x \to \infty} \ln(7x^3 - x^2 + 1) = \infty, \quad \lim_{t \to -\infty} \ln(t^2 - 5t) = \infty \]

\[ \lim_{x \to \infty} \tan^{-1} x = \frac{\pi}{2}, \quad \lim_{x \to -\infty} \tan^{-1} x = -\frac{\pi}{2}, \quad \lim_{x \to \infty} \tan^{-1}(x^3 - 5x + 6) = \frac{\pi}{2}, \quad \lim_{x \to 0^-} \tan^{-1}\left(\frac{1}{x}\right) = -\frac{\pi}{2} \]

Evaluate \( \lim_{x \to 2} (3x^2 + 5x - 1) \).

Solution: Using the sum and constant multiple properties of limits: \[ \lim_{x \to 2} (3x^2 + 5x - 1) = 3 \lim_{x \to 2} x^2 + 5 \lim_{x \to 2} x - 1 = 3(4) + 5(2) -1 = 21 \]

Evaluate \( \lim_{x \to -1} \frac{x^3 + 1}{x+1} \).

Solution: Factor numerator: \(x^3+1=(x+1)(x^2-x+1)\) \[ \lim_{x \to -1} \frac{(x+1)(x^2-x+1)}{x+1} = \lim_{x \to -1} (x^2-x+1) = 3 \]

Evaluate \( \lim_{x \to 0} \frac{\sqrt{x+4} - 2}{x} \).

Solution: Rationalize the numerator: \[ \frac{\sqrt{x+4}-2}{x} \cdot \frac{\sqrt{x+4}+2}{\sqrt{x+4}+2} = \frac{1}{\sqrt{x+4}+2} \implies \lim_{x \to 0} \frac{1}{4} = \frac{1}{4} \]

Evaluate \( \lim_{x \to 1} \frac{x^2-1}{x-1} \).

Solution: Factor numerator: \(x^2-1=(x-1)(x+1)\) \[ \lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} = \lim_{x \to 1} (x+1) = 2 \]

Evaluate \( \lim_{x \to 0} \frac{1-\cos x}{x^2} \) using \(1-\cos x = 2\sin^2(x/2)\).

Solution: \[ \lim_{x \to 0} \frac{2\sin^2(x/2)}{x^2} = \lim_{x \to 0} 2 \left(\frac{\sin(x/2)}{x}\right)^2 = 2\left(\frac{1}{2}\right)^2 = \frac{1}{2} \]

Evaluate \( \lim_{x \to 4} \sqrt{x} \).

Solution: Using the root property: \[ \lim_{x \to 4} \sqrt{x} = \sqrt{\lim_{x \to 4} x} = 2 \]

Evaluate \( \lim_{x \to 0} \frac{\sin 3x}{x} \).

Solution: Factor 3: \[ \lim_{x \to 0} \frac{\sin 3x}{x} = 3 \lim_{x \to 0} \frac{\sin 3x}{3x} = 3(1) = 3 \]

Evaluate \( \lim_{x \to 0} \frac{\tan x}{x} \).

Solution: Standard limit property: \[ \lim_{x \to 0} \frac{\tan x}{x} = 1 \]

Evaluate \( \lim_{x \to 1} \frac{x^4-1}{x-1} \).

Solution: Factor numerator: \(x^4-1 = (x-1)(x^3+x^2+x+1)\) \[ \lim_{x \to 1} \frac{(x-1)(x^3+x^2+x+1)}{x-1} = \lim_{x \to 1} (x^3+x^2+x+1) = 4 \]

Evaluate \( \lim_{x \to 2} \frac{x^2-4}{x^2+4} \).

Solution: Factor numerator: \(x^2-4=(x-2)(x+2)\) \[ \lim_{x \to 2} \frac{(x-2)(x+2)}{x^2+4} = \frac{4}{8} = \frac{1}{2} \]

Evaluate \( \lim_{x \to 0} \frac{e^x-1}{x} \).

Solution: Standard exponential limit: \[ \lim_{x \to 0} \frac{e^x-1}{x} = 1 \]

Evaluate \( \lim_{x \to 0} \frac{\ln(1+x)}{x} \).

Solution: Standard logarithmic limit: \[ \lim_{x \to 0} \frac{\ln(1+x)}{x} = 1 \]

Evaluate \( \lim_{x \to 1} \frac{\sqrt{x}-1}{x-1} \).

Solution: Rationalize numerator: \[ \frac{\sqrt{x}-1}{x-1} \cdot \frac{\sqrt{x}+1}{\sqrt{x}+1} = \frac{1}{\sqrt{x}+1} \implies \lim_{x \to 1} \frac{1}{2} = \frac{1}{2} \]

Evaluate \( \lim_{x \to 0} \frac{\sin x + \tan x}{x} \).

Solution: Split limit: \[ \lim_{x \to 0} \frac{\sin x}{x} + \lim_{x \to 0} \frac{\tan x}{x} = 1 + 1 = 2 \]

Evaluate \( \lim_{x \to 0^+} \ln x \).

Solution: As \(x \to 0^+\), \(\ln x \to -\infty\)

Evaluate \( \lim_{x \to \infty} \frac{3x^2+2}{5x^2-7} \).

Solution: Divide numerator and denominator by \(x^2\): \[ \lim_{x \to \infty} \frac{3+2/x^2}{5-7/x^2} = \frac{3}{5} \]

Evaluate \( \lim_{x \to 0} \frac{1}{x} \).

Solution: Limit does not exist; as \(x \to 0^+\), it \(\to \infty\); as \(x \to 0^-\), it \(\to -\infty\)

Evaluate \( \lim_{x \to 0} \frac{\sin 2x}{\sin 3x} \).

Solution: Rewrite: \[ \lim_{x \to 0} \frac{2x}{3x} \cdot \frac{\sin 2x/(2x)}{\sin 3x/(3x)} = \frac{2}{3} \cdot 1 = \frac{2}{3} \]

Evaluate \( \lim_{x \to 1} (x^5 - 1) \).

Solution: Direct substitution: \[ \lim_{x \to 1} (x^5-1) = 1-1=0 \]

Evaluate \( \lim_{x \to 0} \frac{e^{2x}-1}{\sin x} \).

Solution: Split and use standard limits: \[ \lim_{x \to 0} \frac{e^{2x}-1}{x} \cdot \frac{x}{\sin x} = 2 \cdot 1 = 2 \]