Mister Exam
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- Limit of the function:
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Limit of (-3*e^(4*x)-2*e^(-x)+5*e^(2*x))/(-4*sqrt(1+5*x)+4*cos(3*x)+5*sin(2*x))
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Limit of (-1+log(x)/log(10))/(-1+sqrt(-9+x))
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Limit of (-1+cos(2*x))/(3*x*sin(x))
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Limit of atan(3*x)/asin(2*x)
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Identical expressions
- (eleven + four *(one +n)^ two + eight *n)/(three + four *n^ two + eight *n)
- (11 plus 4 multiply by (1 plus n) squared plus 8 multiply by n) divide by (3 plus 4 multiply by n squared plus 8 multiply by n)
- (eleven plus four multiply by (one plus n) to the power of two plus eight multiply by n) divide by (three plus four multiply by n to the power of two plus eight multiply by n)
- (11+4*(1+n)2+8*n)/(3+4*n2+8*n)
- 11+4*1+n2+8*n/3+4*n2+8*n
- (11+4*(1+n)²+8*n)/(3+4*n²+8*n)
- (11+4*(1+n) to the power of 2+8*n)/(3+4*n to the power of 2+8*n)
- (11+4(1+n)^2+8n)/(3+4n^2+8n)
- (11+4(1+n)2+8n)/(3+4n2+8n)
- 11+41+n2+8n/3+4n2+8n
- 11+41+n^2+8n/3+4n^2+8n
- (11+4*(1+n)^2+8*n) divide by (3+4*n^2+8*n)
-
Similar expressions
- (11+4*(1+n)^2+8*n)/(3-4*n^2+8*n)
- (11+4*(1+n)^2-8*n)/(3+4*n^2+8*n)
- (11+4*(1+n)^2+8*n)/(3+4*n^2-8*n)
- (11+4*(1-n)^2+8*n)/(3+4*n^2+8*n)
- (11-4*(1+n)^2+8*n)/(3+4*n^2+8*n)
Limit of the function (11+4*(1+n)^2+8*n)/(3+4*n^2+8*n)
The solution
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/ 2 \
|11 + 4*(1 + n) + 8*n|
lim |---------------------|
n->oo| 2 |
\ 3 + 4*n + 8*n /
$$\lim_{n \to \infty}\left(\frac{8 n + \left(4 \left(n + 1\right)^{2} + 11\right)}{8 n + \left(4 n^{2} + 3\right)}\right)$$
Limit((11 + 4*(1 + n)^2 + 8*n)/(3 + 4*n^2 + 8*n), n, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{n \to \infty}\left(\frac{8 n + \left(4 \left(n + 1\right)^{2} + 11\right)}{8 n + \left(4 n^{2} + 3\right)}\right)$$
Let's divide numerator and denominator by n^2:
$$\lim_{n \to \infty}\left(\frac{8 n + \left(4 \left(n + 1\right)^{2} + 11\right)}{8 n + \left(4 n^{2} + 3\right)}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{4 + \frac{16}{n} + \frac{15}{n^{2}}}{4 + \frac{8}{n} + \frac{3}{n^{2}}}\right)$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{4 + \frac{16}{n} + \frac{15}{n^{2}}}{4 + \frac{8}{n} + \frac{3}{n^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{15 u^{2} + 16 u + 4}{3 u^{2} + 8 u + 4}\right)$$
=
$$\frac{15 \cdot 0^{2} + 0 \cdot 16 + 4}{3 \cdot 0^{2} + 0 \cdot 8 + 4} = 1$$
The final answer:
$$\lim_{n \to \infty}\left(\frac{8 n + \left(4 \left(n + 1\right)^{2} + 11\right)}{8 n + \left(4 n^{2} + 3\right)}\right) = 1$$
$$\lim_{n \to \infty}\left(\frac{8 n + \left(4 \left(n + 1\right)^{2} + 11\right)}{8 n + \left(4 n^{2} + 3\right)}\right)$$
Let's divide numerator and denominator by n^2:
$$\lim_{n \to \infty}\left(\frac{8 n + \left(4 \left(n + 1\right)^{2} + 11\right)}{8 n + \left(4 n^{2} + 3\right)}\right)$$ =
$$\lim_{n \to \infty}\left(\frac{4 + \frac{16}{n} + \frac{15}{n^{2}}}{4 + \frac{8}{n} + \frac{3}{n^{2}}}\right)$$
Do Replacement
$$u = \frac{1}{n}$$
then
$$\lim_{n \to \infty}\left(\frac{4 + \frac{16}{n} + \frac{15}{n^{2}}}{4 + \frac{8}{n} + \frac{3}{n^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{15 u^{2} + 16 u + 4}{3 u^{2} + 8 u + 4}\right)$$
=
$$\frac{15 \cdot 0^{2} + 0 \cdot 16 + 4}{3 \cdot 0^{2} + 0 \cdot 8 + 4} = 1$$
The final answer:
$$\lim_{n \to \infty}\left(\frac{8 n + \left(4 \left(n + 1\right)^{2} + 11\right)}{8 n + \left(4 n^{2} + 3\right)}\right) = 1$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{n \to \infty}\left(8 n + 4 \left(n + 1\right)^{2} + 11\right) = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty}\left(4 n^{2} + 8 n + 3\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{8 n + \left(4 \left(n + 1\right)^{2} + 11\right)}{8 n + \left(4 n^{2} + 3\right)}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{n \to \infty}\left(\frac{8 n + 4 \left(n + 1\right)^{2} + 11}{4 n^{2} + 8 n + 3}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(8 n + 4 \left(n + 1\right)^{2} + 11\right)}{\frac{d}{d n} \left(4 n^{2} + 8 n + 3\right)}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{8 n + 16}{8 n + 8}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(8 n + 16\right)}{\frac{d}{d n} \left(8 n + 8\right)}\right)$$
=
$$\lim_{n \to \infty} 1$$
=
$$\lim_{n \to \infty} 1$$
=
$$1$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 2 time(s)
oo/oo,
i.e. limit for the numerator is
$$\lim_{n \to \infty}\left(8 n + 4 \left(n + 1\right)^{2} + 11\right) = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty}\left(4 n^{2} + 8 n + 3\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{8 n + \left(4 \left(n + 1\right)^{2} + 11\right)}{8 n + \left(4 n^{2} + 3\right)}\right)$$
=
Let's transform the function under the limit a few
$$\lim_{n \to \infty}\left(\frac{8 n + 4 \left(n + 1\right)^{2} + 11}{4 n^{2} + 8 n + 3}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(8 n + 4 \left(n + 1\right)^{2} + 11\right)}{\frac{d}{d n} \left(4 n^{2} + 8 n + 3\right)}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{8 n + 16}{8 n + 8}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(8 n + 16\right)}{\frac{d}{d n} \left(8 n + 8\right)}\right)$$
=
$$\lim_{n \to \infty} 1$$
=
$$\lim_{n \to \infty} 1$$
=
$$1$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 2 time(s)
The graph
Other limits n→0, -oo, +oo, 1
$$\lim_{n \to \infty}\left(\frac{8 n + \left(4 \left(n + 1\right)^{2} + 11\right)}{8 n + \left(4 n^{2} + 3\right)}\right) = 1$$
$$\lim_{n \to 0^-}\left(\frac{8 n + \left(4 \left(n + 1\right)^{2} + 11\right)}{8 n + \left(4 n^{2} + 3\right)}\right) = 5$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{8 n + \left(4 \left(n + 1\right)^{2} + 11\right)}{8 n + \left(4 n^{2} + 3\right)}\right) = 5$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{8 n + \left(4 \left(n + 1\right)^{2} + 11\right)}{8 n + \left(4 n^{2} + 3\right)}\right) = \frac{7}{3}$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{8 n + \left(4 \left(n + 1\right)^{2} + 11\right)}{8 n + \left(4 n^{2} + 3\right)}\right) = \frac{7}{3}$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{8 n + \left(4 \left(n + 1\right)^{2} + 11\right)}{8 n + \left(4 n^{2} + 3\right)}\right) = 1$$
More at n→-oo
$$\lim_{n \to 0^-}\left(\frac{8 n + \left(4 \left(n + 1\right)^{2} + 11\right)}{8 n + \left(4 n^{2} + 3\right)}\right) = 5$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{8 n + \left(4 \left(n + 1\right)^{2} + 11\right)}{8 n + \left(4 n^{2} + 3\right)}\right) = 5$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{8 n + \left(4 \left(n + 1\right)^{2} + 11\right)}{8 n + \left(4 n^{2} + 3\right)}\right) = \frac{7}{3}$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{8 n + \left(4 \left(n + 1\right)^{2} + 11\right)}{8 n + \left(4 n^{2} + 3\right)}\right) = \frac{7}{3}$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{8 n + \left(4 \left(n + 1\right)^{2} + 11\right)}{8 n + \left(4 n^{2} + 3\right)}\right) = 1$$
More at n→-oo
The graph