Question

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) $lim_{x
ightarrow0}\frac{e^{x}sqrt{x + 1}}{cos(x)+\tan(x)}$ ii) $lim_{x
ightarrow2}ln(sqrt{x + 1})$ iii) $lim_{x
ightarrow4}arcsinleft(\frac{x^{2}-7x + 12}{x^{2}-6x + 8}
ight)$ iv) $lim_{x
ightarrowinfty}cosleft(\frac{pi x^{4}+2x + 1}{3x^{4}+3}
ight)$

Explanation:

i)

Step1: Substitute \(x = 0\)

We know that \(\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x
ightarrow a}f(x)}{\lim_{x
ightarrow a}g(x)}\) (when \(\lim_{x
ightarrow a}g(x)
eq0\)). Here, \(f(x)=e^{x}\sqrt{x + 1}\) and \(g(x)=\cos(x)+\tan(x)\).
When \(x = 0\), \(e^{0}=1\), \(\sqrt{0 + 1}=1\), \(\cos(0)=1\) and \(\tan(0)=0\).
So \(\lim_{x
ightarrow0}\frac{e^{x}\sqrt{x + 1}}{\cos(x)+\tan(x)}=\frac{e^{0}\sqrt{0 + 1}}{\cos(0)+\tan(0)}=\frac{1\times1}{1 + 0}=1\)

Step1: Substitute \(x = 2\)

For the function \(y=\ln(\sqrt{x + 1})\), we use the property \(\lim_{x
ightarrow a}\ln(f(x))=\ln(\lim_{x
ightarrow a}f(x))\) (when \(\lim_{x
ightarrow a}f(x)>0\)).
When \(x = 2\), \(\sqrt{2+1}=\sqrt{3}\).
So \(\lim_{x
ightarrow2}\ln(\sqrt{x + 1})=\ln(\sqrt{2 + 1})=\ln(\sqrt{3})=\frac{1}{2}\ln(3)\)

Step1: Simplify the rational - function inside the arcsin

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

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

\(\lim_{x
ightarrow4}\frac{x^{2}-7x + 12}{x^{2}-6x + 8}=\lim_{x
ightarrow4}\frac{x - 3}{x - 2}=\frac{4-3}{4 - 2}=\frac{1}{2}\)
\(\lim_{x
ightarrow4}\arcsin(\frac{x^{2}-7x + 12}{x^{2}-6x + 8})=\arcsin(\frac{1}{2})=\frac{\pi}{6}\)

Answer:

\(1\)

ii)