Mister Exam
Other calculators:
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How to use it?
- Limit of the function:
-
Limit of ((3+2*x)/(-1+2*x))^(1+x)
-
Limit of (-6-x+2*x^2)/(-2+x)
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Limit of (sqrt(-1+2*x)-(-2+3*x)^(1/5))/(-1+x)
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Limit of (4+2*n^3)/(5+n^2)
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Identical expressions
- (one + three ^(one +n))/(one + three ^n)
- (1 plus 3 to the power of (1 plus n)) divide by (1 plus 3 to the power of n)
- (one plus three to the power of (one plus n)) divide by (one plus three to the power of n)
- (1+3(1+n))/(1+3n)
- 1+31+n/1+3n
- 1+3^1+n/1+3^n
- (1+3^(1+n)) divide by (1+3^n)
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Similar expressions
- (1-3^(1+n))/(1+3^n)
- (1+3^(1-n))/(1+3^n)
- (1+3^(1+n))/(1-3^n)
Limit of the function (1+3^(1+n))/(1+3^n)
The solution
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[src]
/ 1 + n\
|1 + 3 |
lim |----------|
n->oo| n |
\ 1 + 3 /
$$\lim_{n \to \infty}\left(\frac{3^{n + 1} + 1}{3^{n} + 1}\right)$$
Limit((1 + 3^(1 + n))/(1 + 3^n), n, oo, dir='-')
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{n \to \infty}\left(3 \cdot 3^{n} + 1\right) = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty}\left(3^{n} + 1\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{3^{n + 1} + 1}{3^{n} + 1}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(3 \cdot 3^{n} + 1\right)}{\frac{d}{d n} \left(3^{n} + 1\right)}\right)$$
=
$$\lim_{n \to \infty} 3$$
=
$$\lim_{n \to \infty} 3$$
=
$$3$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)
oo/oo,
i.e. limit for the numerator is
$$\lim_{n \to \infty}\left(3 \cdot 3^{n} + 1\right) = \infty$$
and limit for the denominator is
$$\lim_{n \to \infty}\left(3^{n} + 1\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{n \to \infty}\left(\frac{3^{n + 1} + 1}{3^{n} + 1}\right)$$
=
$$\lim_{n \to \infty}\left(\frac{\frac{d}{d n} \left(3 \cdot 3^{n} + 1\right)}{\frac{d}{d n} \left(3^{n} + 1\right)}\right)$$
=
$$\lim_{n \to \infty} 3$$
=
$$\lim_{n \to \infty} 3$$
=
$$3$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)
The graph
Other limits n→0, -oo, +oo, 1
$$\lim_{n \to \infty}\left(\frac{3^{n + 1} + 1}{3^{n} + 1}\right) = 3$$
$$\lim_{n \to 0^-}\left(\frac{3^{n + 1} + 1}{3^{n} + 1}\right) = 2$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{3^{n + 1} + 1}{3^{n} + 1}\right) = 2$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{3^{n + 1} + 1}{3^{n} + 1}\right) = \frac{5}{2}$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{3^{n + 1} + 1}{3^{n} + 1}\right) = \frac{5}{2}$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{3^{n + 1} + 1}{3^{n} + 1}\right) = 1$$
More at n→-oo
$$\lim_{n \to 0^-}\left(\frac{3^{n + 1} + 1}{3^{n} + 1}\right) = 2$$
More at n→0 from the left
$$\lim_{n \to 0^+}\left(\frac{3^{n + 1} + 1}{3^{n} + 1}\right) = 2$$
More at n→0 from the right
$$\lim_{n \to 1^-}\left(\frac{3^{n + 1} + 1}{3^{n} + 1}\right) = \frac{5}{2}$$
More at n→1 from the left
$$\lim_{n \to 1^+}\left(\frac{3^{n + 1} + 1}{3^{n} + 1}\right) = \frac{5}{2}$$
More at n→1 from the right
$$\lim_{n \to -\infty}\left(\frac{3^{n + 1} + 1}{3^{n} + 1}\right) = 1$$
More at n→-oo
The graph