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