Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of 1-cos(3*x)
-
Limit of (-10+6*x^2+7*x)/(-4+x^2)
-
Limit of (7+n)/(5+n)
-
Limit of -3+x*(7-6*x)^x/3
-
Identical expressions
- (three + five *x)/(one +x)
- (3 plus 5 multiply by x) divide by (1 plus x)
- (three plus five multiply by x) divide by (one plus x)
- (3+5x)/(1+x)
- 3+5x/1+x
- (3+5*x) divide by (1+x)
-
Similar expressions
- (3-5*x)/(1+x)
- (-1+3*x^3+5*x)/(1+x^2)
- (-1+x^3+5*x)/(1+x^8)
- (3+5*x)/(1+x^3-4*x)
- (3+5*x)/(1-x)
Limit of the function (3+5*x)/(1+x)
The solution
You have entered
[src]
/3 + 5*x\ lim |-------| x->oo\ 1 + x /
$$\lim_{x \to \infty}\left(\frac{5 x + 3}{x + 1}\right)$$
Limit((3 + 5*x)/(1 + x), x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(\frac{5 x + 3}{x + 1}\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(\frac{5 x + 3}{x + 1}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{5 + \frac{3}{x}}{1 + \frac{1}{x}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{5 + \frac{3}{x}}{1 + \frac{1}{x}}\right) = \lim_{u \to 0^+}\left(\frac{3 u + 5}{u + 1}\right)$$
=
$$\frac{0 \cdot 3 + 5}{1} = 5$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{5 x + 3}{x + 1}\right) = 5$$
$$\lim_{x \to \infty}\left(\frac{5 x + 3}{x + 1}\right)$$
Let's divide numerator and denominator by x:
$$\lim_{x \to \infty}\left(\frac{5 x + 3}{x + 1}\right)$$ =
$$\lim_{x \to \infty}\left(\frac{5 + \frac{3}{x}}{1 + \frac{1}{x}}\right)$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty}\left(\frac{5 + \frac{3}{x}}{1 + \frac{1}{x}}\right) = \lim_{u \to 0^+}\left(\frac{3 u + 5}{u + 1}\right)$$
=
$$\frac{0 \cdot 3 + 5}{1} = 5$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{5 x + 3}{x + 1}\right) = 5$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty}\left(5 x + 3\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(x + 1\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{5 x + 3}{x + 1}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(5 x + 3\right)}{\frac{d}{d x} \left(x + 1\right)}\right)$$
=
$$\lim_{x \to \infty} 5$$
=
$$\lim_{x \to \infty} 5$$
=
$$5$$
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
$$\lim_{x \to \infty}\left(5 x + 3\right) = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(x + 1\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{5 x + 3}{x + 1}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(5 x + 3\right)}{\frac{d}{d x} \left(x + 1\right)}\right)$$
=
$$\lim_{x \to \infty} 5$$
=
$$\lim_{x \to \infty} 5$$
=
$$5$$
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 x→0, -oo, +oo, 1
$$\lim_{x \to \infty}\left(\frac{5 x + 3}{x + 1}\right) = 5$$
$$\lim_{x \to 0^-}\left(\frac{5 x + 3}{x + 1}\right) = 3$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{5 x + 3}{x + 1}\right) = 3$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{5 x + 3}{x + 1}\right) = 4$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{5 x + 3}{x + 1}\right) = 4$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{5 x + 3}{x + 1}\right) = 5$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{5 x + 3}{x + 1}\right) = 3$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{5 x + 3}{x + 1}\right) = 3$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{5 x + 3}{x + 1}\right) = 4$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{5 x + 3}{x + 1}\right) = 4$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{5 x + 3}{x + 1}\right) = 5$$
More at x→-oo
The graph