Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of (-5+7*n)/(2+n)
-
Limit of (-64+4^x)/(-3+x)
-
Limit of (-7+3*x^2+5*x)/(1+x+3*x^2)
-
Limit of (-2+2*x^3+7*x)/(-4-x+3*x^3)
-
Identical expressions
- k/(one +k)
- k divide by (1 plus k)
- k divide by (one plus k)
- k/1+k
- k divide by (1+k)
-
Similar expressions
- k/(1-k)
- ((1+k)/(1+k^2))^(1/3)
Limit of the function k/(1+k)
The solution
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/ k \ lim |-----| k->oo\1 + k/
$$\lim_{k \to \infty}\left(\frac{k}{k + 1}\right)$$
Limit(k/(1 + k), k, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{k \to \infty}\left(\frac{k}{k + 1}\right)$$
Let's divide numerator and denominator by k:
$$\lim_{k \to \infty}\left(\frac{k}{k + 1}\right)$$ =
$$\lim_{k \to \infty} \frac{1}{1 + \frac{1}{k}}$$
Do Replacement
$$u = \frac{1}{k}$$
then
$$\lim_{k \to \infty} \frac{1}{1 + \frac{1}{k}} = \lim_{u \to 0^+} \frac{1}{u + 1}$$
=
$$1^{-1} = 1$$
The final answer:
$$\lim_{k \to \infty}\left(\frac{k}{k + 1}\right) = 1$$
$$\lim_{k \to \infty}\left(\frac{k}{k + 1}\right)$$
Let's divide numerator and denominator by k:
$$\lim_{k \to \infty}\left(\frac{k}{k + 1}\right)$$ =
$$\lim_{k \to \infty} \frac{1}{1 + \frac{1}{k}}$$
Do Replacement
$$u = \frac{1}{k}$$
then
$$\lim_{k \to \infty} \frac{1}{1 + \frac{1}{k}} = \lim_{u \to 0^+} \frac{1}{u + 1}$$
=
$$1^{-1} = 1$$
The final answer:
$$\lim_{k \to \infty}\left(\frac{k}{k + 1}\right) = 1$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{k \to \infty} k = \infty$$
and limit for the denominator is
$$\lim_{k \to \infty}\left(k + 1\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{k \to \infty}\left(\frac{k}{k + 1}\right)$$
=
$$\lim_{k \to \infty}\left(\frac{\frac{d}{d k} k}{\frac{d}{d k} \left(k + 1\right)}\right)$$
=
$$\lim_{k \to \infty} 1$$
=
$$\lim_{k \to \infty} 1$$
=
$$1$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)
oo/oo,
i.e. limit for the numerator is
$$\lim_{k \to \infty} k = \infty$$
and limit for the denominator is
$$\lim_{k \to \infty}\left(k + 1\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{k \to \infty}\left(\frac{k}{k + 1}\right)$$
=
$$\lim_{k \to \infty}\left(\frac{\frac{d}{d k} k}{\frac{d}{d k} \left(k + 1\right)}\right)$$
=
$$\lim_{k \to \infty} 1$$
=
$$\lim_{k \to \infty} 1$$
=
$$1$$
It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 1 time(s)
The graph
Other limits k→0, -oo, +oo, 1
$$\lim_{k \to \infty}\left(\frac{k}{k + 1}\right) = 1$$
$$\lim_{k \to 0^-}\left(\frac{k}{k + 1}\right) = 0$$
More at k→0 from the left
$$\lim_{k \to 0^+}\left(\frac{k}{k + 1}\right) = 0$$
More at k→0 from the right
$$\lim_{k \to 1^-}\left(\frac{k}{k + 1}\right) = \frac{1}{2}$$
More at k→1 from the left
$$\lim_{k \to 1^+}\left(\frac{k}{k + 1}\right) = \frac{1}{2}$$
More at k→1 from the right
$$\lim_{k \to -\infty}\left(\frac{k}{k + 1}\right) = 1$$
More at k→-oo
$$\lim_{k \to 0^-}\left(\frac{k}{k + 1}\right) = 0$$
More at k→0 from the left
$$\lim_{k \to 0^+}\left(\frac{k}{k + 1}\right) = 0$$
More at k→0 from the right
$$\lim_{k \to 1^-}\left(\frac{k}{k + 1}\right) = \frac{1}{2}$$
More at k→1 from the left
$$\lim_{k \to 1^+}\left(\frac{k}{k + 1}\right) = \frac{1}{2}$$
More at k→1 from the right
$$\lim_{k \to -\infty}\left(\frac{k}{k + 1}\right) = 1$$
More at k→-oo
The graph