The limits doesn't exist, but it does have the correct form so it might be an infinite limit.

Rewriting 1/ x 2 − 1/ x 3 as an equivalent fractional expression ( x − 1)/ x 3, the numerator approaches −1, and the denominator approaches 0 through positive values as x approaches 0 from the right; hence, the function decreases without bound and . This means $$x^2> 25$$.

= \frac{-6+1}{-3+3} We know the limit doesn't exist.

% %

\\ Three Ways to Find Limits Involving Infinity. \displaystyle\lim_{x\to 8}\,\frac{x^2-7x-8}{(x-8)^2} In general, a fractional function will have an infinite limit if the limit of the denominator is zero and the limit … The numerator approaches 9, so it is positive.

Review your understanding of infinite limits with some challenge problems. $$, $$

= \frac{-6} 0

Here’s how that works: If the degrees of the two polynomials are equal, there’s a horizontal asymptote at the number you get when you divide the coefficient of the highest power term in the numerator by the coefficient of the highest power term in the denominator. Since x is approaching 6 from the left, we know $$x<6$$, so $$x^2 < 36$$, so the denominator will be negative. (1) lim x->2 (x - 2)/(x 2 - x - 2) You have to pull a positive out of the radicand (as always), so you pull out negative 4x because when x is negative (which it is as it approaches negative infinity), –4x is positive.

In this case, the coefficients of x 2 are 6 in the numerator and 1 in the denominator. One sided Limits, Next % The numerator approaches 45, so it will be positive. $$\displaystyle \lim_{x\to8^-}\,\frac{x+1}{x-8} = -\infty$$, $$\displaystyle \lim_{x\to8^+}\,\frac{x+1}{x-8} = \infty$$. $$. % At some point in your calculus life, you’ll be asked to find a limit at infinity. The limit doesn't exist, but it has the $$\frac n 0$$ form so it might be an infinite limit.

$$.

% However, it has the correct form so it might be an infinite limit. $$. As the denominator shrinks to 0, the function will become infinitely large. The comment form collects the name and email you enter, and the content, to allow us keep track of the comments placed on the website. Examples and interactive practice problems, explained and worked out step by step As the denominator approaches zero, the function will become infinitely large. But in case it helps, here are the steps you can use: First let $y = (1/x)^{(1/x)}$. &= \infty \quad \cmark \end{align*} \] The second line in the solution shows that the function approaches $\dfrac{x}{3}$  as x grows large, matching what the graph shows. Result: $$\displaystyle \lim_{x\to3^-}\,\frac{x+2}{x+3} = -\infty$$, Result: $$\displaystyle \lim_{x\to3^+}\,\frac{x+2}{x+3} = \infty$$. So, $$\displaystyle \lim_{x\to 4^-}\,\frac{x+4}{x-4} = -\infty$$, So, $$\displaystyle \lim_{x\to 4^+}\,\frac{x+4}{x-4} = \infty$$.

Since $$x$$ is approaching 8 from the left, the denominator is negative.

The limit does not exist.

The sign of the infinite limit is determined by the sign of the quotient of the numerator and the denominator at values close to the number that the independent variable is approaching. More specifically, we know that the limit is either $\infty$ or $-\infty$. So the denominator will be positive. Taken all together, the statements tell us, $$\displaystyle \lim_{x\to-3^-}\,\frac{x-5}{x+3} = \infty$$, $$\displaystyle \lim_{x\to-3^+}\,\frac{x-5}{x+3} = -\infty$$. To determine the limit at infinity we need only look at the term with the highest power in the numerator, and the term with the highest power in the denominator.

Take all together, these statements mean both one-sided limits are infinite in the positive direction. %

\displaystyle \lim_{x\to7}\,\frac{x^2-4}{(x-7)^2} = \infty In general, a fractional function will have an infinite limit if the limit of the denominator is zero and the limit of the numerator is not zero. Try factoring the denominator so the one-sided limits are easier to analyze. %
It may seem strange, but infinity minus infinity does not equal 0. \displaystyle \lim_{x\to 2} \,\frac{2x^2-14}{x^2-4} %

= \displaystyle\lim_{x\to\frac 1 2}\,\frac{6x-9}{(2x-1)^2} Then \begin{align*} \ln y &= \ln(1/x)^{(1/x)} \\[8px] &= (1/x) \ln (1/x) \\[8px] &= -\frac{\ln (x)}{x} \end{align*} where in the second line we made use of the fact that $\ln a^b = b \ln a,$ and in the third line $\ln(1/x) = -\ln x.$ From there, if you take $ \displaystyle{\lim_{x \to \infty }}$ on both sides, on the right you have $\dfrac{\infty}{\infty}$ and so you can apply L’Hôpital’s Rule. \displaystyle \lim_{x\to 6}\,\frac{x^2 - 5x - 36}{x^2-36} This means $$x^2 < 25$$.

$$\\ \displaystyle \lim_{x\to2^-}\, \frac{2x^2-14}{x^2-4} = \infty \\$$, $$\displaystyle \lim_{x\to2^+}\, \frac{2x^2-14}{x^2-4} = -\infty$$. The numerator approaches -6, so it will be negative.

\displaystyle \lim_{x\to -\frac 3 2}\,\frac{4x+1}{2x+3} This is a classic L’Hôpital’s Rule question, which this page doesn’t address. A word of caution: Do not evaluate the limits individually and subtract because ±∞ are not real numbers.

The numerator approaches 5, so it will be positive.

The numerator approaches -5, so it will be negative. Infinite Limits--When Limits Do not exist because the function becomes infinitey large. \displaystyle \lim_{x\to -\frac 3 2}\,\frac{8x^2+14x+3}{4x^2 + 12x + 9}

When this occurs, the function is said to have an infinite limit; hence, you write .
The limit at infinity does not exist because the function continually oscillates between -1 and 1 forever as x grows and Grows. The denominator is a perfect square, so as $$x$$ approaches $$\frac 1 2$$, the denominator will always be positive. In both cases, the denominator is approaching 0, so the function will become infinitely large. For problems 7 & 8 find all the vertical asymptotes of the given function. Infinite limits of functions are found by looking at the end behavior of functions. The limit doesn't exist, but it has the correct form so it might be an infinite limit.

We use cookies to provide you the best possible experience on our website.

As x approaches 2 from the left, the numerator approaches 5, and the denominator approaches 0 through negative values; hence, the function decreases without bound and . Note that had you plugged in infinity in the original problem, you would have.

He also does extensive one-on-one tutoring.