Limit of the function y+x^2

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     /     2\
 lim \y + x /
x->oo

\lim_{x \to \infty}\left(x^{2} + y\right)

Limit(y + x^2, x, oo, dir='-')

Detail solution

Let's take the limit

\lim_{x \to \infty}\left(x^{2} + y\right)

Let's divide numerator and denominator by x^2:

\lim_{x \to \infty}\left(x^{2} + y\right)

\lim_{x \to \infty}\left(\frac{1 + \frac{y}{x^{2}}}{\frac{1}{x^{2}}}\right)

Do Replacement

u = \frac{1}{x}

then

\lim_{x \to \infty}\left(\frac{1 + \frac{y}{x^{2}}}{\frac{1}{x^{2}}}\right) = \lim_{u \to 0^+}\left(\frac{u^{2} y + 1}{u^{2}}\right)

\frac{0^{2} y + 1}{0} = \infty

The final answer:

\lim_{x \to \infty}\left(x^{2} + y\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

Rapid solution [src]

oo

\infty

Expand and simplify

Other limits x→0, -oo, +oo, 1

\lim_{x \to \infty}\left(x^{2} + y\right) = \infty

\lim_{x \to 0^-}\left(x^{2} + y\right) = y

\lim_{x \to 0^+}\left(x^{2} + y\right) = y

\lim_{x \to 1^-}\left(x^{2} + y\right) = y + 1

\lim_{x \to 1^+}\left(x^{2} + y\right) = y + 1

\lim_{x \to -\infty}\left(x^{2} + y\right) = \infty