Question

1. evaluate the limits

a) \\( \lim _ { x \
ightarrow \infty } \frac { e ^ { x } + 5 x } { x ^ { 2 } + 5 } \\)
b) \\( \lim _ { x \
ightarrow 0 } \frac { \sin ( 3 x ) - 3 x } { x ^ { 3 } } \\)
c) \\( \lim _ { x \
ightarrow \infty } x \sin ( 2 / x ) \\)
d) \\( \lim _ { x \
ightarrow \infty } ( e ^ { 2 x } + 3 ) ^ { 3 / x } \\)
e) \\( \lim _ { x \
ightarrow \infty } \left( 1 + \frac { 3 } { x } \
ight) ^ { 2 x } \\)

Explanation:

Step1: Analyze the limit $\lim_{x

ightarrow\infty}\frac{e^{x}+5x}{x^{2}+5}$
As \(x
ightarrow\infty\), the exponential function \(e^{x}\) grows faster than any polynomial function. We can use L - H rule (since \(\lim_{x
ightarrow\infty}\frac{e^{x}+5x}{x^{2}+5}\) is of the form \(\frac{\infty}{\infty}\)).
Differentiate numerator and denominator: \(\lim_{x
ightarrow\infty}\frac{e^{x}+5}{2x}\) (still \(\frac{\infty}{\infty}\)).
Differentiate again: \(\lim_{x
ightarrow\infty}\frac{e^{x}}{2}=\infty\)

Step2: Analyze the limit \(\lim_{x

ightarrow0}\frac{\sin(3x)-3x}{x^{3}}\)
Use the Maclaurin series \(\sin t=t-\frac{t^{3}}{6}+\frac{t^{5}}{120}-\cdots\). Let \(t = 3x\), then \(\sin(3x)=3x-\frac{(3x)^{3}}{6}+\frac{(3x)^{5}}{120}-\cdots\)
\(\lim_{x
ightarrow0}\frac{3x-\frac{27x^{3}}{6}-3x}{x^{3}}=\lim_{x
ightarrow0}\frac{-\frac{9x^{3}}{2}}{x^{3}}=-\frac{9}{2}\)

Step3: Analyze the limit \(\lim_{x

ightarrow\infty}x\sin(\frac{2}{x})\)
Let \(t=\frac{1}{x}\), as \(x
ightarrow\infty\), \(t
ightarrow0\). Then the limit becomes \(\lim_{t
ightarrow0}\frac{\sin(2t)}{t}\)
Using the fact that \(\lim_{u
ightarrow0}\frac{\sin u}{u} = 1\), with \(u = 2t\), we have \(\lim_{t
ightarrow0}\frac{\sin(2t)}{t}=2\lim_{t
ightarrow0}\frac{\sin(2t)}{2t}=2\)

Step4: Analyze the limit \(\lim_{x

ightarrow\infty}(e^{2x}+3)^{\frac{3}{x}}\)
Take the natural logarithm: \(y=(e^{2x}+3)^{\frac{3}{x}}\), \(\ln y=\frac{3\ln(e^{2x}+3)}{x}\)
As \(x
ightarrow\infty\), \(\lim_{x
ightarrow\infty}\frac{3\ln(e^{2x}+3)}{x}\) (L - H rule, \(\frac{\infty}{\infty}\) form)
Differentiate numerator and denominator: \(\lim_{x
ightarrow\infty}\frac{3\frac{2e^{2x}}{e^{2x}+3}}{1}=6\)
Since \(\lim_{x
ightarrow\infty}\ln y = 6\), then \(\lim_{x
ightarrow\infty}y=e^{6}\)

Step5: Analyze the limit \(\lim_{x

ightarrow\infty}(1 + \frac{3}{x})^{2x}\)
Use the formula \(\lim_{n
ightarrow\infty}(1+\frac{a}{n})^{n}=e^{a}\). Let \(n=x\) and \(a = 6\) (since \((1+\frac{3}{x})^{2x}=[(1+\frac{3}{x})^{x}]^{2}\))
\(\lim_{x
ightarrow\infty}(1+\frac{3}{x})^{2x}=e^{6}\)

Answer:

a) \(\infty\)
b) \(-\frac{9}{2}\)
c) \(2\)
d) \(e^{6}\)
e) \(e^{6}\)