[SOLVED] Calculating limit of a sequence

**mollymcf2009** · March 16th 2009, 10:47 AM

Can someone help me with a good strategy for finding the limit of a sequence? I am just not understanding the best way to solve these. What I have been trying to do to first take the limit. If i got an indeterminate, I changed it to a function and took the derivative, but I'm not sure that is the way I need to evaluate all of these. Any help is greatly appreciated!

1) $a_n = 1 - (0.1)^n $

$\lim_{n\rightarrow \infty}$ $a_n = ?$

2) $a_n = \frac{n^3}{3n + 1}$

$\lim_{n\rightarrow \infty}$ $a_n = ?$

3) $a_n = e^{7/(n+9)}$

$ \lim_{n\rightarrow \infty}$ $a_n = ?$

4) $a_n = \frac{3^{n+1}}{5^n}$

$\lim_{n\rightarrow \infty}$ $a_n = ?$

5) $a_n = tan (\frac{6n\pi}{4 + 24n})$

$\lim_{n\rightarrow \infty}$ $a_n = ?$

6) $a_n = n^2e^{-5n}$

$\lim_{n\rightarrow \infty}$ $a_n = ?$

**Chris L T521** · March 16th 2009, 10:59 AM

Originally Posted by mollymcf2009

Can someone help me with a good strategy for finding the limit of a sequence? I am just not understanding the best way to solve these. What I have been trying to do to first take the limit. If i got an indeterminate, I changed it to a function and took the derivative, but I'm not sure that is the way I need to evaluate all of these. Any help is greatly appreciated!

1) $a_n = 1 - (0.1)^n $

$\lim_{n\rightarrow \infty}$ $a_n = ?$

Hint: $\lim a^n=0$ when $\left|a\right|<1$

2) $a_n = \frac{n^3}{3n + 1}$

$\lim_{n\rightarrow \infty}$ $a_n = ?$

Hint: Divide the numerator and denominator through by $n^3$ to get $\lim\frac{1}{\frac{3}{n^2}+\frac{1}{n^3}}$

Can you take it from here?

3) $a_n = e^{7/(n+9)}$

$ \lim_{n\rightarrow \infty}$ $a_n = ?$

$\lim e^{\frac{7}{n+9}}=e^{\lim\frac{7}{n+9}}$

Can you take it from here?

4) $a_n = \frac{3^{n+1}}{5^n}$

$\lim_{n\rightarrow \infty}$ $a_n = ?$

$\lim\frac{3^{n+1}}{5^n}=3\lim\frac{3^n}{5^n}=3\lim \left(\frac{3}{5}\right)^n$ . Now see my hint for #1

5) $a_n = tan (\frac{6n\pi}{4 + 24n})$

$\lim_{n\rightarrow \infty}$ $a_n = ?$

$\lim\tan\left(\frac{6n\pi}{24n+4}\right)=\tan\left (\lim\frac{6\pi n}{24n+4}\right)$

Can you take it from here?

6) $a_n = n^2e^{-5n}$

$\lim_{n\rightarrow \infty}$ $a_n = ?$

$\lim n^2e^{-5n}=\lim\frac{n^2}{e^{5n}}$

Can you take it from here?

(Note that $\lim$ is analogous with $\lim_{n\to\infty}$ )

**mollymcf2009** · March 16th 2009, 11:47 AM

Originally Posted by mollymcf2009

can someone help me with a good strategy for finding the limit of a sequence? I am just not understanding the best way to solve these. What i have been trying to do to first take the limit. If i got an indeterminate, i changed it to a function and took the derivative, but i'm not sure that is the way i need to evaluate all of these. Any help is greatly appreciated!

1) $a_n = 1 - (0.1)^n $

$\lim_{n\rightarrow \infty}$ $a_n = ?$ 1

2) $a_n = \frac{n^3}{3n + 1}$

$\lim_{n\rightarrow \infty}$ $a_n = ?$ diverges

3) $a_n = e^{7/(n+9)}$

$ \lim_{n\rightarrow \infty}$ $a_n = ?$ 1

4) $a_n = \frac{3^{n+1}}{5^n}$

$\lim_{n\rightarrow \infty}$ $a_n = ?$ 0

5) $a_n = tan (\frac{6n\pi}{4 + 24n})$

$\lim_{n\rightarrow \infty}$ $a_n = ?$ 1

6) $a_n = n^2e^{-5n}$

$\lim_{n\rightarrow \infty}$ $a_n = ?$ 0

correct answers are in red above for future reference

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