Mister Exam

Limit of the function 1+x/2

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The solution

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     /    x\
 lim |1 + -|
x->oo\    2/
limx(x2+1)\lim_{x \to \infty}\left(\frac{x}{2} + 1\right)
Limit(1 + x/2, x, oo, dir='-')
Detail solution
Let's take the limit
limx(x2+1)\lim_{x \to \infty}\left(\frac{x}{2} + 1\right)
Let's divide numerator and denominator by x:
limx(x2+1)\lim_{x \to \infty}\left(\frac{x}{2} + 1\right) =
limx(12+1x1x)\lim_{x \to \infty}\left(\frac{\frac{1}{2} + \frac{1}{x}}{\frac{1}{x}}\right)
Do Replacement
u=1xu = \frac{1}{x}
then
limx(12+1x1x)=limu0+(u+12u)\lim_{x \to \infty}\left(\frac{\frac{1}{2} + \frac{1}{x}}{\frac{1}{x}}\right) = \lim_{u \to 0^+}\left(\frac{u + \frac{1}{2}}{u}\right)
=
102=\frac{1}{0 \cdot 2} = \infty

The final answer:
limx(x2+1)=\lim_{x \to \infty}\left(\frac{x}{2} + 1\right) = \infty
Lopital's rule
There is no sense to apply Lopital's rule to this function since there is no indeterminateness of 0/0 or oo/oo type
The graph
02468-8-6-4-2-1010-1010
Other limits x→0, -oo, +oo, 1
limx(x2+1)=\lim_{x \to \infty}\left(\frac{x}{2} + 1\right) = \infty
limx0(x2+1)=1\lim_{x \to 0^-}\left(\frac{x}{2} + 1\right) = 1
More at x→0 from the left
limx0+(x2+1)=1\lim_{x \to 0^+}\left(\frac{x}{2} + 1\right) = 1
More at x→0 from the right
limx1(x2+1)=32\lim_{x \to 1^-}\left(\frac{x}{2} + 1\right) = \frac{3}{2}
More at x→1 from the left
limx1+(x2+1)=32\lim_{x \to 1^+}\left(\frac{x}{2} + 1\right) = \frac{3}{2}
More at x→1 from the right
limx(x2+1)=\lim_{x \to -\infty}\left(\frac{x}{2} + 1\right) = -\infty
More at x→-oo
Rapid solution [src]
oo
\infty
The graph
Limit of the function 1+x/2