Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of (x^2-x)/(-7+2*x^2+5*x)
-
Limit of x^2/(4+x)
-
Limit of (-20+x^2-x)/(-4+x^2-3*x)
-
Limit of (6+x^2-5*x)/(2-x^2)
- Graphing y =:
- x^2/(4+x)
-
Identical expressions
- x^ two /(four +x)
- x squared divide by (4 plus x)
- x to the power of two divide by (four plus x)
- x2/(4+x)
- x2/4+x
- x²/(4+x)
- x to the power of 2/(4+x)
- x^2/4+x
- x^2 divide by (4+x)
-
Similar expressions
- x^2/(4-x)
- (2-3*x+5*x^2)/(4+x^2+3*x)
Limit of the function x^2/(4+x)
The solution
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lim |-----|
x->oo\4 + x/
$$\lim_{x \to \infty}\left(\frac{x^{2}}{x + 4}\right)$$
Limit(x^2/(4 + x), x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(\frac{x^{2}}{x + 4}\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to \infty}\left(\frac{x^{2}}{x + 4}\right)$$ =
$$\lim_{x \to \infty} \frac{1}{\frac{1}{x} + \frac{4}{x^{2}}}$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty} \frac{1}{\frac{1}{x} + \frac{4}{x^{2}}} = \lim_{u \to 0^+} \frac{1}{4 u^{2} + u}$$
=
$$\frac{1}{4 \cdot 0^{2}} = \infty$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{x^{2}}{x + 4}\right) = \infty$$
$$\lim_{x \to \infty}\left(\frac{x^{2}}{x + 4}\right)$$
Let's divide numerator and denominator by x^2:
$$\lim_{x \to \infty}\left(\frac{x^{2}}{x + 4}\right)$$ =
$$\lim_{x \to \infty} \frac{1}{\frac{1}{x} + \frac{4}{x^{2}}}$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty} \frac{1}{\frac{1}{x} + \frac{4}{x^{2}}} = \lim_{u \to 0^+} \frac{1}{4 u^{2} + u}$$
=
$$\frac{1}{4 \cdot 0^{2}} = \infty$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{x^{2}}{x + 4}\right) = \infty$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty} x^{2} = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(x + 4\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{x^{2}}{x + 4}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} x^{2}}{\frac{d}{d x} \left(x + 4\right)}\right)$$
=
$$\lim_{x \to \infty}\left(2 x\right)$$
=
$$\lim_{x \to \infty}\left(2 x\right)$$
=
$$\infty$$
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
$$\lim_{x \to \infty} x^{2} = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(x + 4\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{x^{2}}{x + 4}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} x^{2}}{\frac{d}{d x} \left(x + 4\right)}\right)$$
=
$$\lim_{x \to \infty}\left(2 x\right)$$
=
$$\lim_{x \to \infty}\left(2 x\right)$$
=
$$\infty$$
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
$$\lim_{x \to \infty}\left(\frac{x^{2}}{x + 4}\right) = \infty$$
$$\lim_{x \to 0^-}\left(\frac{x^{2}}{x + 4}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x^{2}}{x + 4}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x^{2}}{x + 4}\right) = \frac{1}{5}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x^{2}}{x + 4}\right) = \frac{1}{5}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x^{2}}{x + 4}\right) = -\infty$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{x^{2}}{x + 4}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x^{2}}{x + 4}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x^{2}}{x + 4}\right) = \frac{1}{5}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x^{2}}{x + 4}\right) = \frac{1}{5}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x^{2}}{x + 4}\right) = -\infty$$
More at x→-oo
The graph