Mister Exam
Other calculators:
-
How to use it?
- Limit of the function:
-
Limit of x/sqrt(1+x^2)
- Limit of x+y
-
Limit of (-36+x^2)/(-6+x)
-
Limit of (9-x+2*x^2)/(5-x)
- Graphing y =:
- x/sqrt(1+x^2)
- Integral of d{x}:
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x/sqrt(1+x^2)
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Identical expressions
- x/sqrt(one +x^ two)
- x divide by square root of (1 plus x squared )
- x divide by square root of (one plus x to the power of two)
- x/√(1+x^2)
- x/sqrt(1+x2)
- x/sqrt1+x2
- x/sqrt(1+x²)
- x/sqrt(1+x to the power of 2)
- x/sqrt1+x^2
- x divide by sqrt(1+x^2)
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Similar expressions
- x/sqrt(1-x^2)
- (1+4*x)/sqrt(1+x^2)
Limit of the function x/sqrt(1+x^2)
The solution
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/ x \
lim |-----------|
x->-oo| ________|
| / 2 |
\\/ 1 + x /
$$\lim_{x \to -\infty}\left(\frac{x}{\sqrt{x^{2} + 1}}\right)$$
Limit(x/sqrt(1 + x^2), x, -oo)
Lopital's rule
We have indeterminateness of type
i.e. limit for the numerator is
$$\lim_{x \to -\infty} x = -\infty$$
and limit for the denominator is
$$\lim_{x \to -\infty} \sqrt{x^{2} + 1} = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to -\infty}\left(\frac{x}{\sqrt{x^{2} + 1}}\right)$$
=
$$\lim_{x \to -\infty}\left(\frac{\frac{d}{d x} x}{\frac{d}{d x} \sqrt{x^{2} + 1}}\right)$$
=
$$\lim_{x \to -\infty}\left(\frac{\sqrt{x^{2} + 1}}{x}\right)$$
=
$$\lim_{x \to -\infty}\left(\frac{\sqrt{x^{2} + 1}}{x}\right)$$
=
$$-1$$
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
$$\lim_{x \to -\infty} x = -\infty$$
and limit for the denominator is
$$\lim_{x \to -\infty} \sqrt{x^{2} + 1} = \infty$$
Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.
$$\lim_{x \to -\infty}\left(\frac{x}{\sqrt{x^{2} + 1}}\right)$$
=
$$\lim_{x \to -\infty}\left(\frac{\frac{d}{d x} x}{\frac{d}{d x} \sqrt{x^{2} + 1}}\right)$$
=
$$\lim_{x \to -\infty}\left(\frac{\sqrt{x^{2} + 1}}{x}\right)$$
=
$$\lim_{x \to -\infty}\left(\frac{\sqrt{x^{2} + 1}}{x}\right)$$
=
$$-1$$
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 x→0, -oo, +oo, 1
$$\lim_{x \to -\infty}\left(\frac{x}{\sqrt{x^{2} + 1}}\right) = -1$$
$$\lim_{x \to \infty}\left(\frac{x}{\sqrt{x^{2} + 1}}\right) = 1$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{x}{\sqrt{x^{2} + 1}}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x}{\sqrt{x^{2} + 1}}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x}{\sqrt{x^{2} + 1}}\right) = \frac{\sqrt{2}}{2}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x}{\sqrt{x^{2} + 1}}\right) = \frac{\sqrt{2}}{2}$$
More at x→1 from the right
$$\lim_{x \to \infty}\left(\frac{x}{\sqrt{x^{2} + 1}}\right) = 1$$
More at x→oo
$$\lim_{x \to 0^-}\left(\frac{x}{\sqrt{x^{2} + 1}}\right) = 0$$
More at x→0 from the left
$$\lim_{x \to 0^+}\left(\frac{x}{\sqrt{x^{2} + 1}}\right) = 0$$
More at x→0 from the right
$$\lim_{x \to 1^-}\left(\frac{x}{\sqrt{x^{2} + 1}}\right) = \frac{\sqrt{2}}{2}$$
More at x→1 from the left
$$\lim_{x \to 1^+}\left(\frac{x}{\sqrt{x^{2} + 1}}\right) = \frac{\sqrt{2}}{2}$$
More at x→1 from the right
The graph