Question

1. calculate the values of the following limits. simplify your answers as much as possible, but leave them as exact values. explain how you arrived at your answer.

see examples 1.8.1, 1.8.2 in chad’s calculus notes for similar types of problems
i) $lim_{x \to 0}\frac{e^{x}sqrt{x + 1}}{cos(x)+\tan(x)}$ ii) $lim_{x \to 2}ln(sqrt{x + 1})$ iii) $lim_{x \to 4}arcsin(\frac{x^{2}-7x + 12}{x^{2}-6x + 8})$ iv) $lim_{x \to infty}cos(\frac{pi x^{4}+2x + 1}{3x^{4}+3})$

Explanation:

i)

Step1: Substitute \(x = 0\)

Substitute \(x=0\) into \(\frac{e^{x}\sqrt{x + 1}}{\cos(x)+\tan(x)}\). We know that \(e^{0}=1\), \(\sqrt{0 + 1}=1\), \(\cos(0)=1\) and \(\tan(0)=0\).
\[
\frac{e^{0}\sqrt{0 + 1}}{\cos(0)+\tan(0)}=\frac{1\times1}{1 + 0}
\]

Step2: Simplify the expression

\[
\frac{1\times1}{1+0}=1
\]

Step1: Substitute \(x = 2\)

Substitute \(x = 2\) into \(\ln(\sqrt{x + 1})\). First, find the value of \(\sqrt{x+1}\) when \(x = 2\), \(\sqrt{2+1}=\sqrt{3}\).
Then we have \(\ln(\sqrt{3})\)

Step1: Simplify the rational - function inside the arcsin

First, factor the numerator \(x^{2}-7x + 12=(x - 3)(x - 4)\) and the denominator \(x^{2}-6x + 8=(x - 2)(x - 4)\). So \(\frac{x^{2}-7x + 12}{x^{2}-6x + 8}=\frac{(x - 3)(x - 4)}{(x - 2)(x - 4)}=\frac{x - 3}{x - 2}\) for \(x
eq4\).

Step2: Substitute \(x = 4\) into the simplified function

\(\lim_{x
ightarrow4}\arcsin(\frac{x^{2}-7x + 12}{x^{2}-6x + 8})=\lim_{x
ightarrow4}\arcsin(\frac{x - 3}{x - 2})\). Substitute \(x = 4\) into \(\frac{x - 3}{x - 2}\), we get \(\frac{4-3}{4 - 2}=\frac{1}{2}\).
Then \(\arcsin(\frac{1}{2})\)

Step3: Evaluate the arcsin function

We know that \(\arcsin(\frac{1}{2})=\frac{\pi}{6}\)

Answer:

\(1\)

ii)