5. evaluate the limits\na) \\( \\lim _ { x \\rightarrow \\infty } \\frac { e ^ { x } + 5 x } { x ^ { 2 } + 5…

5. evaluate the limits\na) \\( \\lim _ { x \\rightarrow \\infty } \\frac { e ^ { x } + 5 x } { x ^ { 2 } + 5 } \\)\nb) \\( \\lim _ { x \\rightarrow 0 } \\frac { \\sin ( 3 x ) - 3 x } { x ^ { 3 } } \\)\nc) \\( \\lim _ { x \\rightarrow \\infty } x \\sin ( 2 / x ) \\)\nd) \\( \\lim _ { x \\rightarrow \\infty } ( e ^ { 2 x } + 3 ) ^ { 3 / x } \\)\ne) \\( \\lim _ { x \\rightarrow \\infty } \\left( 1 + \\frac { 3 } { x } \\right) ^ { 2 x } \\)

5. evaluate the limits\na) \\( \\lim _ { x \\rightarrow \\infty } \\frac { e ^ { x } + 5 x } { x ^ { 2 } + 5 } \\)\nb) \\( \\lim _ { x \\rightarrow 0 } \\frac { \\sin ( 3 x ) - 3 x } { x ^ { 3 } } \\)\nc) \\( \\lim _ { x \\rightarrow \\infty } x \\sin ( 2 / x ) \\)\nd) \\( \\lim _ { x \\rightarrow \\infty } ( e ^ { 2 x } + 3 ) ^ { 3 / x } \\)\ne) \\( \\lim _ { x \\rightarrow \\infty } \\left( 1 + \\frac { 3 } { x } \\right) ^ { 2 x } \\)

Answer

Explanation:

Step1: Analyze the limit $\lim_{x\rightarrow\infty}\frac{e^{x}+5x}{x^{2}+5}$

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

Step2: Analyze the limit (\lim_{x\rightarrow0}\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\rightarrow0}\frac{3x-\frac{27x^{3}}{6}-3x}{x^{3}}=\lim_{x\rightarrow0}\frac{-\frac{9x^{3}}{2}}{x^{3}}=-\frac{9}{2})

Step3: Analyze the limit (\lim_{x\rightarrow\infty}x\sin(\frac{2}{x}))

Let (t=\frac{1}{x}), as (x\rightarrow\infty), (t\rightarrow0). Then the limit becomes (\lim_{t\rightarrow0}\frac{\sin(2t)}{t}) Using the fact that (\lim_{u\rightarrow0}\frac{\sin u}{u} = 1), with (u = 2t), we have (\lim_{t\rightarrow0}\frac{\sin(2t)}{t}=2\lim_{t\rightarrow0}\frac{\sin(2t)}{2t}=2)

Step4: Analyze the limit (\lim_{x\rightarrow\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\rightarrow\infty), (\lim_{x\rightarrow\infty}\frac{3\ln(e^{2x}+3)}{x}) (L - H rule, (\frac{\infty}{\infty}) form) Differentiate numerator and denominator: (\lim_{x\rightarrow\infty}\frac{3\frac{2e^{2x}}{e^{2x}+3}}{1}=6) Since (\lim_{x\rightarrow\infty}\ln y = 6), then (\lim_{x\rightarrow\infty}y=e^{6})

Step5: Analyze the limit (\lim_{x\rightarrow\infty}(1 + \frac{3}{x})^{2x})

Use the formula (\lim_{n\rightarrow\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\rightarrow\infty}(1+\frac{3}{x})^{2x}=e^{6})

Answer:

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