Mister Exam
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How to use it?
- Limit of the function:
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Limit of -sin(x)+tan(x)
-
Limit of cos(x)*tan(5*x)
-
Limit of ((-1+4*x)/(3+4*x))^(2+3*x)
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Limit of 1/(n*log(n))
- Integral of d{x}:
-
x^4/(1+x^2)
-
Identical expressions
- x^ four /(one +x^ two)
- x to the power of 4 divide by (1 plus x squared )
- x to the power of four divide by (one plus x to the power of two)
- x4/(1+x2)
- x4/1+x2
- x⁴/(1+x²)
- x to the power of 4/(1+x to the power of 2)
- x^4/1+x^2
- x^4 divide by (1+x^2)
-
Similar expressions
- x^4/(1-x^2)
Limit of the function x^4/(1+x^2)
The solution
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/ 4 \
| x |
lim |------|
x->oo| 2|
\1 + x /
$$\lim_{x \to \infty}\left(\frac{x^{4}}{x^{2} + 1}\right)$$
Limit(x^4/(1 + x^2), x, oo, dir='-')
Detail solution
Let's take the limit
$$\lim_{x \to \infty}\left(\frac{x^{4}}{x^{2} + 1}\right)$$
Let's divide numerator and denominator by x^4:
$$\lim_{x \to \infty}\left(\frac{x^{4}}{x^{2} + 1}\right)$$ =
$$\lim_{x \to \infty} \frac{1}{\frac{1}{x^{2}} + \frac{1}{x^{4}}}$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty} \frac{1}{\frac{1}{x^{2}} + \frac{1}{x^{4}}} = \lim_{u \to 0^+} \frac{1}{u^{4} + u^{2}}$$
=
$$\frac{1}{0^{2} + 0^{4}} = \infty$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{x^{4}}{x^{2} + 1}\right) = \infty$$
$$\lim_{x \to \infty}\left(\frac{x^{4}}{x^{2} + 1}\right)$$
Let's divide numerator and denominator by x^4:
$$\lim_{x \to \infty}\left(\frac{x^{4}}{x^{2} + 1}\right)$$ =
$$\lim_{x \to \infty} \frac{1}{\frac{1}{x^{2}} + \frac{1}{x^{4}}}$$
Do Replacement
$$u = \frac{1}{x}$$
then
$$\lim_{x \to \infty} \frac{1}{\frac{1}{x^{2}} + \frac{1}{x^{4}}} = \lim_{u \to 0^+} \frac{1}{u^{4} + u^{2}}$$
=
$$\frac{1}{0^{2} + 0^{4}} = \infty$$
The final answer:
$$\lim_{x \to \infty}\left(\frac{x^{4}}{x^{2} + 1}\right) = \infty$$
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to \infty} x^{4} = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(x^{2} + 1\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{x^{4}}{x^{2} + 1}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} x^{4}}{\frac{d}{d x} \left(x^{2} + 1\right)}\right)$$
=
$$\lim_{x \to \infty}\left(2 x^{2}\right)$$
=
$$\lim_{x \to \infty}\left(2 x^{2}\right)$$
=
$$\infty$$
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
$$\lim_{x \to \infty} x^{4} = \infty$$
and limit for the denominator is
$$\lim_{x \to \infty}\left(x^{2} + 1\right) = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to \infty}\left(\frac{x^{4}}{x^{2} + 1}\right)$$
=
$$\lim_{x \to \infty}\left(\frac{\frac{d}{d x} x^{4}}{\frac{d}{d x} \left(x^{2} + 1\right)}\right)$$
=
$$\lim_{x \to \infty}\left(2 x^{2}\right)$$
=
$$\lim_{x \to \infty}\left(2 x^{2}\right)$$
=
$$\infty$$
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
$$\lim_{x \to \infty}\left(\frac{x^{4}}{x^{2} + 1}\right) = \infty$$
$$\lim_{x \to 0^-}\left(\frac{x^{4}}{x^{2} + 1}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x^{4}}{x^{2} + 1}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x^{4}}{x^{2} + 1}\right) = \frac{1}{2}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x^{4}}{x^{2} + 1}\right) = \frac{1}{2}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x^{4}}{x^{2} + 1}\right) = \infty$$
More at x→-oo
$$\lim_{x \to 0^-}\left(\frac{x^{4}}{x^{2} + 1}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x^{4}}{x^{2} + 1}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x^{4}}{x^{2} + 1}\right) = \frac{1}{2}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x^{4}}{x^{2} + 1}\right) = \frac{1}{2}$$
More at x→1 from the right
$$\lim_{x \to -\infty}\left(\frac{x^{4}}{x^{2} + 1}\right) = \infty$$
More at x→-oo
The graph