There are several approaches used to find limits. Here, we summarize the different strategies, and their advantages and disadvantages.

Many limits may be evaluated by substitution. The necessary requirement for this approach to work is that the function is continuous at the point where the limit is being evaluated.

- $\lim\limits_{x\to 5} (x^2-3x+2)=5^2-3(5)+2=12$

**Caution:** When evaluating, if the expression $\sqrt{0}$
is encountered, it is also necessary to determine whether the result is
valid for a two-sided limit, or for a particular one-sided limit, or
possibly not valid at all.

- $\lim\limits_{x\to 6+}\sqrt{6-x}$, when evaluated by substitution, gives $\sqrt{6-6}=\sqrt{0}=0$. However, this limit actually
**does not exist**, because the domain of the function $f(x)=\sqrt{6-x}$ is $(-\infty,6]$, and therefore $x$ cannot approach 6 from the right along this function. The only limit of this function that would exist as $x$ approaches 6 is a limit from the left.

When evaluation produces the form $\dfrac{0}{0}$, quite frequently there is a common factor in the numerator and denominator of the function which can be cancelled.

- $\lim\limits_{x\to 4}\dfrac{x^2-10x+24}{x^2-x-12}$, when evaluated by substitution, produces the form $\dfrac{0}{0}$. So we proceed as follows:

$\lim\limits_{x\to 4} \dfrac{x^2-10x+24}{x^2-x-12}=\lim\limits_{x\to 4}\dfrac{(x-6)(x-4)}{(x-4)(x+3)}= \lim\limits_{x\to 4}\dfrac{x-6}{x+3}=\dfrac{4-6}{4+3}=-\dfrac{2}{7}$.

Sometimes when the form $\dfrac{0}{0}$ occurs, algebraic manipulation is required before the factors are apparent. With compound fractions, a common denominator should be obtained.

- $\lim\limits_{x\to 2}\dfrac{\frac{1}{x}-\frac12}{x-2}=\lim\limits_{x\to 2} \dfrac{\left(\frac{2-x}{2x}\right)}{\left(\frac{x-2}{1}\right)}=\lim\limits_{x\to 2} \left(\dfrac{x-2}{-2x}\right)\left(\dfrac{1}{x-2}\right)=\lim\limits_{x\to 2} \dfrac{-1}{2x}=-\dfrac14$

When the form $\dfrac{0}{0}$ occurs and square roots are present, the numerator and denominator should be multiplied by the conjugate of the expression containing the square root.

- $\lim\limits_{x\to 3}\dfrac{\sqrt{x^2+16}-5}{x^2-9}=\lim\limits_{x\to 3}
\dfrac{\left(\sqrt{x^2+16}-5\right)\left(\sqrt{x^2+16}+5\right)}
{(x^2-9)\left(\sqrt{x^2+16}+5\right)}=\lim\limits_{x\to 3}\dfrac{x^2+16-25}
{(x^2-9)\left(\sqrt{x^2+16}+5\right)}$

$\qquad\qquad =\lim\limits_{x\to 3}\dfrac{x^2-9} {(x^2-9)\left(\sqrt{x^2+16}+5\right)}=\lim\limits_{x\to 3}\dfrac{1}{\left(\sqrt{x^2+16}+5\right)} =\dfrac{1}{10}$

When the form $\dfrac{0}{0}$ occurs and absolute values are present, it is possible that only a one-sided limit will exist. The absolute value will need to be rewritten. In this example, the argument of the absolute value is negative when approaching from the left, so a factor of negative one appears in the solution.

- $\lim\limits_{x\to 7-}\dfrac{(x-4)|x-7|}{x-7}=\lim\limits_{x\to 7-}\dfrac{(x-4)(-1)(x-7)} {x-7}=\lim\limits_{x\to 7-}(-1)(x-4)=(-1)(7-4)=-3$

If evaluation produces the form $\dfrac{0}{0}$ and the sine function is present, the special case $\lim\limits{\theta\to 0}\dfrac{\sin\theta}{\theta}=1$ may be needed. Note that this approach is valid only when the argument of the sine function is approaching zero (not infinity).

- $\lim\limits_{x\to 0}\dfrac{4\sin 5x}{3x}=\lim\limits_{x\to 0} \dfrac{\frac53 (4\sin 5x)}{\frac53 (3x)}=\lim\limits_{x\to 0}\dfrac{20}{3} \left(\dfrac{\sin 5x}{5x}\right)=\dfrac{20}{3}(1)=\dfrac{20}{3}$

When the form $\dfrac{0}{0}$ appears with a cosine function, the same special case is needed, along with some trigonometric identities.

- $\lim\limits_{x\to 0}\dfrac{1-\cos 4x}{3x}=\lim\limits_{x\to 0}\dfrac{(1-\cos 4x)(1+\cos 4x)}
{3x(1+\cos 4x)}=\lim\limits_{x\to 0}\dfrac{1-\cos^2 4x}{3x(1+\cos 4x)}=\lim\limits_{x\to 0}
\dfrac{\frac43 \sin^2 4x}{\frac43 (3x)(1+\cos 4x)}$

$\qquad\qquad =\lim\limits_{x\to 0}\left(\dfrac43 \dfrac{\sin 4x}{4x}\dfrac{\sin 4x}{1+\cos 4x}\right) =\dfrac43 (1)\left(\dfrac02\right)=0$

Sometimes evaluation produces the form $\dfrac{\infty}{\infty}$. When this occurs, the limit may be of a rational function as $x$ approaches infinity. This can be solved by multiplying the numerator and denominator by the reciprocal of the highest power of $x$ appearing in the function, together with the special case $\lim\limits_{x\to\infty}\dfrac{1}{x}=0$.

- $\lim\limits_{x\to\infty}\dfrac{5x^3-4x^2+11}{6x^3-9x+7}=\lim\limits_{x\to\infty}\dfrac {(5x^3-4x^2+11)\left(\frac{1}{x^3}\right)}{(6x^3-9x+7)\left(\frac{1}{x^3}\right)}= \lim\limits_{x\to\infty}\dfrac{5-\frac{4}{x}+\frac{11}{x^3}}{6-\frac{9}{x^2}+\frac{7}{x^3}} =\dfrac{5-0+0}{6-0+0}=\dfrac56$

Another case that produces the form $\dfrac{\infty}{\infty}$ involves exponential functions. These require the special case $\lim\limits_{x\to\infty}b^{-x}=0$, valid for bases $b>1$.

- $\lim\limits_{x\to\infty}\dfrac{5+e^{3x}}{4-2e^{3x}}=\lim\limits_{x\to\infty}\dfrac {(5+e^{3x})e^{-3x}}{(4-2e^{3x})e^{-3x}}=\lim\limits_{x\to\infty}\dfrac{5e^{-3x}+1}{4e^{-3x}-2} =\dfrac{0+1}{0-2}=-\dfrac12$

Some functions are quite similar to known functions, but need sandwiching to handle them properly.

- $\lim\limits_{x\to\infty}\dfrac{1}{x-4}$ is similar to $\lim\limits_{x\to\infty}\dfrac{1}{x}$, so we would expect a similar result. Substitution won't produce that result, but we can sandwich it. Unfortunately, whenever $x>4$, then $\dfrac{1}{x-4} > \dfrac{1}{x}$, so the function $\dfrac{1}{x}$ will not work as the upper layer of the sandwich. But $\dfrac{2}{x}$ will work, since the inequality $\dfrac{2}{x} > \dfrac{1}{x-4}$ will be true whenever $x>8$ (which you can confirm by solving the inequality). So we have $0\le\dfrac{1}{x-4}\le\dfrac{2}{x}$, and taking limits gives $\lim\limits_{x\to\infty}0\le\lim\limits_{x\to\infty}\dfrac{1}{x-4} \le\lim\limits_{x\to\infty}\dfrac{2}{x}$. But both the leftmost and rightmost expressions have limit zero, so we have $0\le\lim\limits_{x\to\infty}\dfrac{1}{x-4}\le 0$, and thus $\lim\limits_{x\to\infty}\dfrac{1}{x-4}=0$.

Some functions cannot be evaluated at their limit, and algebraic manipulation will not simplify the expression. In some of these cases, the Sandwich Theorem may be usable.

- $\lim\limits_{x\to 0}\left(x^2\sin\dfrac{1}{x}\right)$ cannot be simplified into another form. But because the sine function has a limited range, this function can be sandwiched. We first note that $-1\le\sin\dfrac{1}{x}\le 1$ is true for every nonzero real number $x$. Multiplying both sides by $x^2$ produces $-x^2\le x^2\sin\dfrac{1}{x}\le x^2$, and this does not change the direction of the inequalities since $x^2$ is never negative. Then we take limits of each expression to obtain $\lim\limits_{x\to 0}(-x^2)\le \lim\limits_{x\to 0}\left(x^2\sin\dfrac{1}{x}\right)\le \lim\limits_{x\to 0}(x^2)$, and after evaluating the leftmost and rightmost limits we have $0\le \lim\limits_{x\to 0}\left(x^2\sin\dfrac{1}{x}\right)\le 0$. So our limit has been sandwiched between two functions which have the same limit, and we are forced to the conclusion that $\lim\limits_{x\to 0}\left(x^2\sin\dfrac{1}{x}\right)= 0$.

In the above example, note that as $x$ approaches zero, the argument of the sine function approaches infinity, so the special case $\lim\limits_{\theta\to 0}\dfrac{\sin\theta}{\theta}=1$ will not apply.

Several forms are considered indeterminate forms, for which a limit may or may not exist. Besides the most common situations of $\dfrac00$ and $\dfrac{\infty}{\infty}$, there are many other situations which an algebraic approach should be pursued. These include $0\cdot\infty$, $\infty-\infty$, $0^0$, $1^\infty$, and $\infty^0$. Some of these will be studied later in a section on L'Hopital's Rule.

If evaluation produces the form $\dfrac10$, then the limit involves an infinite discontinuity. The specific result can usually be determined by analyzing the signs of the factors involved.

- $\lim\limits_{x\to 2+}\dfrac{x-6}{x-2}$, when evaluated by substitution, produces $\dfrac{-4}{0}$, which is the same form as $\dfrac10$. The numerator is clearly negative near $x=2$ (as we learned when we evaluated). As $x$ approaches 2 from the right, we have $x>2$, which implies $x-2>0$, so the denominator is positive. Thus, we have the quotient of a negative and a positive, which is negative. Therefore, $\lim\limits_{x\to 2+}\dfrac{x-6}{x-2}=-\infty$.

It should be noted that functions will often (but not always) have different infinite limits from the right and from the left. When this occurs, the two-sided limit does not exist.

Graphs may be used to help you understand the situation, but they should generally not be used for justification of results. The technology (both hardware and software) that produces graphical results generally has resolution issues at some scale, and therefore cannot be considered conclusive by itself. But graphs can be very useful when needing to identify a certain type of discontinuity.

Numerical approximations can also be useful in understanding a situation, but they will almost never give conclusive results. Just as graphs do, tables of numerical values also suffer from the "resolution" problem. But when a function has a finite limit, and efforts at simplifying the function fail, a numerical approximation may be the best available result.