# Teach Yourself Limits in 8 Hours: Part 2

After the definitions and basic examples in Part 1, we now focus on the rules of evaluation of limits which will be highly useful in solving various limit problems. We will postpone the proofs of these rules to the last post in the series to avoid any distraction.

### Rules of Limits

In the following rules we assume that the functions described are defined in a certain neighborhood of $a$ except possibly at point $a$. All the relations between the functions (if any) also hold in this neighborhood of point $a$ (except possibly at point $a$).

1) If $f(x) \leq g(x)$ and both $\lim_{x \to a}f(x)$ and $\lim_{x \to a}g(x)$ exist then we have $$\lim_{x \to a}f(x) \leq \lim_{x \to a}g(x)$$ Even if $f(x) < g(x)$ then also we get $$\lim_{x \to a}f(x) \leq \lim_{x \to a}g(x)$$ so that taking limits always weakens the inequalities.

2) $\displaystyle \lim_{x \to a}k = k$ where $k$ is a constant (i.e. not dependent on $x$).

3) $\displaystyle \lim_{x \to a}\{f(x) \pm g(x)\} = \lim_{x \to a}f(x) \pm \lim_{x \to a}g(x)$ provided both the limits on the right side exist.

4) $\displaystyle \lim_{x \to a}f(x) \cdot g(x) = \lim_{x \to a}f(x) \cdot \lim_{x \to a}g(x)$ provided both the limits on the right side exist.

5) $\displaystyle \lim_{x \to a}\frac{f(x)}{g(x)} = \dfrac{{\displaystyle \lim_{x \to a}f(x)}}{{\displaystyle \lim_{x \to a}g(x)}}$ provided both the limits on the right side exist and $\lim_{x \to a}g(x) \neq 0$

6) If $f(x) \leq g(x) \leq h(x)$ and $\displaystyle \lim_{x \to a}f(x) = \lim_{x \to a}h(x) = L$ then $\displaystyle \lim_{x \to a}g(x) = L$. The inequalities can be strict without any change in conclusion. This result is also known as Sandwich Theorem or Squeeze Theorem.

7) If $\displaystyle \lim_{x \to a}g(x) = b$ and $\lim_{x \to b}f(x) = L$ and $g(x) \neq b$ in a certain neighborhood of $a$ (except possibly at $a$) then $\displaystyle \lim_{x \to a}f\{g(x)\} = L$. This is known as the rule of substitution.

There are similar rules for the scenarios when $x \to \infty$ and $x \to -\infty$ which the reader can easily formulate for himself. These simple rules are almost always provided in all introductory calculus texts but are never used with explicit knowledge by beginners. Most students underestimate the power of these rules and prefer to use high level techniques (to be discussed in next post).

We now provide some simple applications of these rules. Since $\lim_{x \to a}x = a$ it follows that if $$f(x) = a_{0}x^{n} + a_{1}x^{n - 1} + \cdots + a_{n - 1}x + a_{n}$$ is a polynomial then $\lim_{x \to a}f(x) = f(a)$ so that in case of polynomials we can simply put $x = a$ and evaluate the limit. This technique works only because of the rules above and may not work for functions other than polynomials.

Going further using rule 5) we can see that the technique of substituting $x = a$ will work for rational functions (ratio of two polynomials) also provided that after the substitution the denominator does not vanish.

Again we can easily establish (directly using the definition) that $\lim_{x \to 0}|x| = 0$. And from basic trigonometry we have the inequality $|\sin x| \leq |x|$ for $|x| \leq \pi/2$. Thus we see that $-|x| \leq \sin x \leq |x|$ and by Sandwich theorem we get the result $$\lim_{x \to 0}\sin x = 0$$ Next we can proceed as follows: \begin{align} \lim_{x \to 0}\cos x &= \lim_{x \to 0} 1 - 2\sin^{2}(x/2)\notag\\ &= 1 - 2\lim_{x \to 0}\sin^{2}(x/2)\notag\\ &= 1 - 2\lim_{y \to 0}\sin^{2}y\text{ (putting }y = x/2\text{)}\notag\\ &= 1 - 2\lim_{y \to 0}\sin y\lim_{y \to 0}\sin y\notag\\ &= 1 - 2\cdot 0\cdot 0 = 1\notag \end{align} Using these limits we can see that \begin{align} \lim_{x \to a}\sin x &= \lim_{h \to 0}\sin(a + h)\text{ (putting }x = a + h)\notag\\ &= \lim_{h \to 0}\sin a\cos h + \cos a\sin h\notag\\ &= \sin a\lim_{h \to 0}\cos h + \cos a\lim_{h \to 0}\sin h\notag\\ &= \sin a\cdot 1 + \cos a\cdot 0 = \sin a\notag \end{align} and in the same way we can prove that $\lim_{x \to a}\cos x = \cos a$.

It now follows that if we have to evaluate the limit of a rational function in $\sin x, \cos x$ we can use the technique of substituting $x = a$ provided this substitution does not lead to a zero denominator.

### Standard Limits

We have seen that for certain functions evaluation of a limit is equivalent to substituting $x = a$ in the given expression for the function provided such a substitution does not lead to a zero denominator. But most of the limit problems one encounters in calculus often lead to zero denominators and for such problems we make use of the following standard limits:

1) If $x^{n}$ is defined in a neighborhood of $a$ (and also at point $a$) then $$\lim_{x \to a}\frac{x^{n} - a^{n}}{x - a} = na^{n - 1}$$ provided $a^{n - 1}$ is also defined.

The above rule helps in calculating various limits related to algebraic functions. In some simpler cases it might to be easier to do algebraic manipulations (like rationalization to remove radicals) rather than using this formula. From the above formula it also follows that \begin{align} \lim_{x \to a}x^{n} &= \lim_{x \to a}(x - a) \cdot\frac{x^{n} - a^{n}}{x - a} + a^{n}\notag\\ &= (a - a)\cdot na^{n - 1} + a^{n} = a^{n}\notag \end{align} provided that $a^{n}$ is defined.

2) $\displaystyle \lim_{x \to 0}\frac{\sin x}{x} = 1$

This is the fundamental limit used to evaluate limits involving trigonometric functions. Normally this is used in conjuction with various trigonometric identities and manipulation based on them. For example \begin{align} \lim_{x \to 0}\frac{1 - \cos x}{x^{2}} &= \lim_{x \to 0}\dfrac{2\sin^{2}(x/2)}{x^{2}}\notag\\ &= 2\lim_{x \to 0}\dfrac{\sin^{2}(x/2)}{(x/2)^{2}}\frac{(x/2)^{2}}{x^{2}}\notag\\ &= 2\cdot 1\cdot \frac{1}{4} = \frac{1}{2}\notag \end{align} 3) $\displaystyle \lim_{x \to 0}\frac{e^{x} - 1}{x} = 1$

This result handles limits involving exponential functions and has to be used in conjuction with the properties of exponential function. Using this formula we can derive another important result: \begin{align} \lim_{x \to 0}\frac{a^{x} - 1}{x} &= \lim_{x \to 0}\frac{e^{x\log a} - 1}{x}\notag\\ &= \lim_{x \to 0}\frac{e^{x\log a} - 1}{x\log a}\frac{x\log a}{x}\notag\\ &= 1\cdot \log a = \log a\notag \end{align} 4) $\displaystyle \lim_{x \to 0}\frac{\log (1 + x)}{x} = 1$

5) $\displaystyle \lim_{x \to \infty}\frac{\log x}{x^{a}} = 0$ for all $a > 0$. It has a counterpart namely that
$\displaystyle \lim_{x \to \infty}\frac{x^{a}}{e^{x}} = 0$

6) $e^{x} \to \infty$ as $x \to \infty$ and $\log x \to \infty$ as $x \to \infty$

Using the above results we can see that \begin{align} \lim_{x \to a}e^{x} &= \lim_{h \to 0}e^{a + h}\notag\\ &= \lim_{h \to 0}e^{a}\cdot e^{h}\notag\\ &= e^{a}\lim_{h \to 0}\left\{h\cdot\left(\frac{e^{h} - 1}{h}\right) + 1\right\}\notag\\ &= e^{a}(0\cdot 1 + 1) = e^{a}\notag \end{align} and if $a > 0$ then \begin{align} \lim_{x \to a}\log x &= \lim_{h \to 0}\log(a + h)\notag\\ &= \lim_{h \to 0}\log\left\{a\left(1 + \frac{h}{a}\right)\right\}\notag\\ &= \lim_{h \to 0}\left\{\log a + \log\left(1 + \frac{h}{a}\right)\right\}\notag\\ &= \log a + \lim_{h \to 0}\frac{h}{a}\cdot\dfrac{\log\left(1 + \dfrac{h}{a}\right)}{\dfrac{h}{a}}\notag\\ &= \log a + 0\cdot 1 = \log a\notag \end{align} From the above we can see that subtitution $x = a$ is allowed in case of logarithmic and exponential functions also provided that we take care of the fact that logarithm is defined for positive numbers only. Using the previous results of this kind we formulate our thumb rule as follows:

To evaluate the limit of an expression (consisting of trigonometric, logarithmic, exponential and algebraical functions and arithmetical operations) when $x \to a$, it is OK to substitute $x = a$ provided that such substitution does not lead to an undefined expression (like zero denominator, square root of negative number, $\log$ of zero or a negative number, etc). Also in case of exponential expression it is important that either the base or the exponent must be a constant. To handle expressions like $\{f(x)\}^{g(x)}$ it is important to recast them in the form $\exp\{g(x)\log(f(x))\}$ and then perform substitution.

However as will be seen in the examples to follow, most usual limit problems will involve expressions where substitution leads to undefined expressions. Hence we have to do certain manipulations to transform the expression into a form which allows us to take advantage of the above standard limits (note that all standard limits lead to undefined expression if we use direct substitution). This manipulation requires a kind of skill which comes with practice.

### Handling Infinity ($\infty$)

Before we proceed to demonstrate the effectiveness of standard limit formulas it is necessary to clarify some misconceptions about the use of $\infty$ in limit problems. This particular aspect of limit problem dealing with $\infty$ is the most confusing to beginners and I have seen ample examples of confusion on MSE. Here are some simple thumb rules to follow and thereby steer clear of any confusion:
• $\infty$ is not a number and hence can't be used with common operations like $+, -, \times, /, =$ like real numbers.
• $\infty$ has a meaning in certain contexts only because of special definitions which give it a meaning in that context.
• Rules for substituting $x = a$ mentioned in last subsection apply only for $x \to a$ scenario and not for $x \to \pm \infty$ scenarios
• If $f(x) \to \infty$ (or $f(x) \to -\infty$) as $x \to a$ then $\displaystyle \lim_{x \to a}\dfrac{1}{f(x)} = 0$
• However, if $\displaystyle \lim_{x \to a}f(x) = 0$ then it does not necessarily follow that $1/f(x) \to \infty$ (or $1/f(x) \to -\infty$) as $x \to a$. If $f(x)$ maintains a constant sign, say positive (or negative) as $x \to a$ then $1/f(x) \to \infty$ (or $1/f(x) \to -\infty$) as $x \to a$. If $f(x)$ does not maintain a constant sign as $x \to a$, then $1/f(x)$ oscillates infinitely as $x \to a$.
• In many cases the algebraical manipulation becomes simpler if the limit $x \to \infty$ (or $x \to -\infty$) is replaced by $\lim_{h \to 0+}$ (or $\lim_{h \to 0-}$) by putting $x = 1/h$.
• If $f(x) \to \infty$ as $x \to a$ and $k > 0$ then $kf(x) \to \infty$ as $x \to a$ and if $k < 0$ then $kf(x) \to -\infty$ as $x \to a$.
• If $f(x) \to \infty$ and $g(x) \to \infty$ as $x \to a$ then $f(x) + g(x) \to \infty$ as $x \to a$.
• If $f(x) \to \infty$ and $g(x) \to \infty$ as $x \to a$ then $f(x) \cdot g(x) \to \infty$ as $x \to a$.
The last three rules of thumb are important and they have counterparts for $-\infty$ version which the reader can formulate himself. One may ask why we don't have similar rules for $f(x) - g(x)$ and $f(x)/g(x)$. Just to clarify, each of the above thumb rules we have mentioned can be proved rigorously (and will be done in the last post of this series) using definition of limit and these proofs are not applicable to these cases. To provide an example let $f(x) = 2/x, g(x) = 1/x$ then $f(x) - g(x) = 1/x \to \infty$ as $x \to 0+$. But if $f(x) = 1/x, g(x) = 1/x$ then $f(x) - g(x) = 0$ and this tends to $0$ when $x \to 0+$.

Another question which might come to mind about handling of $kf(x)$ is what happens when $k = 0$. Clearly when $k = 0$ then $kf(x) = 0$ and thus $\lim_{x \to a}kf(x) = 0$ irrespective of the nature and behavior of $f(x)$. On the other hand if we have $f(x) \to 0$ as $x \to a$ and $g(x) \to \infty$ as $x \to a$ then we can't say anything about limit of $f(x)g(x)$ as $x \to a$. We will see that in such cases often the manipulation of the expression allows us to use some standard limits.

### Examples

I will provide some examples which may seem tough but can be handled easily with the above rules and standard limits.

1) $\displaystyle \lim_{x \to 0}\frac{1 - \cos(1 - \cos x)}{x^{4}}$
We can proceed as follows: \begin{align} \lim_{x \to 0}\frac{1 - \cos(1 - \cos x)}{x^{4}} &= \lim_{x \to 0}\frac{1 - \cos(2\sin^{2}(x/2))}{x^{4}}\notag\\ &= \lim_{x \to 0}\frac{2\sin^{2}(\sin^{2}(x/2))}{x^{4}}\notag\\ &= 2\lim_{x \to 0}\dfrac{\sin^{2}(\sin^{2}(x/2))}{\sin^{4}(x/2)}\frac{\sin^{4}(x/2)}{x^{4}}\notag\\ &= 2\lim_{x \to 0}\left(\dfrac{\sin(\sin^{2}(x/2))}{\sin^{2}(x/2)}\right)^{2}\left(\dfrac{\sin(x/2)}{(x/2)}\right)^{4}\frac{(x/2)^{4}}{x^{4}}\notag\\ &= 2\cdot 1^{2}\cdot 1^{4}\cdot\frac{1}{16} = \frac{1}{8}\notag\end{align} This seemingly tough example is taken from G. H. Hardy's "A Course of Pure Mathematics".

2) $\displaystyle \lim_{x \to a+}\frac{\log(x - a)}{\log(e^{x} - e^{a})}$
Clearly we have \begin{align} L &= \lim_{x \to a+}\frac{\log(x - a)}{\log(e^{x} - e^{a})}\notag\\ &= \lim_{h \to 0+}\frac{\log h}{\log(e^{a + h} - e^{a})}\notag\\ &= \lim_{h \to 0+}\frac{\log h}{\log\{e^{a}(e^{h} - 1)\}}\notag\\ &= \lim_{h \to 0+}\frac{\log h}{a + \log(e^{h} - 1)}\notag\\ &= \lim_{h \to 0+}\dfrac{\log h}{a + \log\left(h\cdot\dfrac{e^{h} - 1}{h}\right)}\notag\\ &= \lim_{h \to 0+}\dfrac{\log h}{a + \log h + \log\left(\dfrac{e^{h} - 1}{h}\right)}\notag\\ &= \lim_{h \to 0+}\dfrac{1}{\dfrac{a}{\log h} + 1 + \dfrac{1}{\log h}\cdot\log\left(\dfrac{e^{h} - 1}{h}\right)}\notag\\ &= \frac{1}{0 + 1 + 0\cdot 0} = 1\notag \end{align} Here if $h \to 0+$ and $y = 1/h$ then $\log h = \log (1/y) = -\log y$ and $y \to \infty$ so that $\log y \to \infty$. It follows that $\log h \to -\infty$ as $h \to 0+$ and hence $1/\log h \to 0$ as $h \to 0+$.

This problem is taken from MSE.
3) $\displaystyle \lim_{x \to 0}\left(\frac{1}{\log(x + \sqrt{1 + x^{2}})} - \frac{1}{\log(1 + x)}\right)$
I presented the following solution on MSE: \begin{align} L &= \lim_{x \to 0}\left(\frac{1}{\log(x + \sqrt{1 + x^{2}})} - \frac{1}{\log(1 + x)}\right)\notag\\ &= \lim_{x \to 0}\frac{\log(1 + x) - \log(x + \sqrt{1 + x^{2}})}{\log(x + \sqrt{1 + x^{2}})\log(1 + x)}\notag\\ &= \lim_{x \to 0}\frac{\log(1 + x) - \log(x + \sqrt{1 + x^{2}})}{\log(1 - 1 + x + \sqrt{1 + x^{2}})\log(1 + x)}\notag\\ &= \lim_{x \to 0}\dfrac{\log(1 + x) - \log(x + \sqrt{1 + x^{2}})}{{\displaystyle \left(x + \sqrt{1 + x^{2}} - 1\right)\dfrac{\log\left\{1 + \left(x + \sqrt{1 + x^{2}} - 1\right)\right\}}{x + \sqrt{1 + x^{2}} - 1}\cdot x\cdot\dfrac{\log(1 + x)}{x}}}\notag\\ &= \lim_{x \to 0}\dfrac{\log(1 + x) - \log(x + \sqrt{1 + x^{2}})}{{\displaystyle \left(x + \sqrt{1 + x^{2}} - 1\right)1\cdot x\cdot 1}}\notag\\ &= \lim_{x \to 0}\dfrac{\log(1 + x) - \log(x + \sqrt{1 + x^{2}})}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\notag\\ &= \lim_{x \to 0}\dfrac{\log(1 + x) - \log\left(\dfrac{\sqrt{1 + x^{2}} - x}{\sqrt{1 + x^{2}} - x}\cdot\left(x + \sqrt{1 + x^{2}}\right)\right)}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\notag\\ &= \lim_{x \to 0}\dfrac{\log(1 + x) + \log\left(\sqrt{1 + x^{2}} - x\right)}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\notag\\ &= \lim_{x \to 0}\dfrac{\log\left\{(1 + x)\left(\sqrt{1 + x^{2}} - x\right)\right\}}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\notag\\ &= \lim_{x \to 0}\dfrac{\log\left\{\sqrt{1 + x^{2}} - x + x\sqrt{1 + x^{2}} - x^{2}\right\}}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\notag\\ &= \lim_{x \to 0}\dfrac{\log\left\{1 - 1 + \sqrt{1 + x^{2}} - x + x\sqrt{1 + x^{2}} - x^{2}\right\}}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\notag\\ &= \lim_{x \to 0}\dfrac{\log\left\{1 + \left(\sqrt{1 + x^{2}} - x + x\sqrt{1 + x^{2}} - x^{2} - 1\right)\right\}}{\left(\sqrt{1 + x^{2}} - x + x\sqrt{1 + x^{2}} - x^{2} - 1\right)}\notag\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\times\dfrac{\left(\sqrt{1 + x^{2}} - x + x\sqrt{1 + x^{2}} - x^{2} - 1\right)}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\notag\\ &= \lim_{x \to 0}\dfrac{\left(\sqrt{1 + x^{2}} - x + x\sqrt{1 + x^{2}} - x^{2} - 1\right)}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\notag\\ &= \lim_{x \to 0}\dfrac{\left(x^{2} + x\sqrt{1 + x^{2}} - x + \sqrt{1 + x^{2}}  - 2x^{2} - 1\right)}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\notag\\ &= \lim_{x \to 0} 1 + \dfrac{\sqrt{1 + x^{2}}  - 2x^{2} - 1}{x \left(x + \sqrt{1 + x^{2}} - 1\right)}\notag\\ &= 1 + \lim_{x \to 0}\dfrac{1 + x^{2} - (2x^{2} + 1)^{2}}{x \left(1 + x^{2} - (1 - x)^{2}\right)}\frac{\sqrt{1 + x^{2}} + 1 - x}{\sqrt{1 + x^{2}} + 2x^{2} + 1}\notag\\ &= 1 + \lim_{x \to 0}\dfrac{-3x^{2} - 4x^{4}}{2x^{2}}\cdot\frac{2}{2} = 1 - \frac{3}{2} = -\frac{1}{2}\notag \end{align} If we observe the steps above carefully we will find that the logarithmic limit has been used smartly to get rid of the logarithm function itself and the expression is transformed into an algebraic one. The calculation of algebraic limits is far simpler than those containing logarithms and exponentials and in this particular example we have used rationalization to get rid of radicals.

### Misuse of Rules of Limits

Before concluding this post I would like to warn against the misuse of rules of limits which deal with operations of $+, -, \times, /$. Each of these rules involves two functions and limits of both the functions must exist for the rule to be applicable. A common misuse is to apply the rules even when limit of one function does not exist.

Another very popular misuse of the rules is the following (and there are many variations on it):
Let $\lim_{x \to a}h(x) = A$ and $$\lim_{x \to a}\frac{f(x)}{g(x) + h(x)} = \lim_{x \to a}\frac{f(x)}{g(x) + A}\textbf{ (wrong)}$$ This step is wrong!! Rules of limits allow us to split one limit operation on a complex expression into multiple limit operations applied to each smaller part of the expression connected by $+, -, \times, /$. Unless the split of the limit operation is not done completely till the smaller parts of expression one can't replace smaller parts by their limits. In the above example the valid step is like this: $$\lim_{x \to a}\frac{f(x)}{g(x) + h(x)} = \dfrac{{\displaystyle \lim_{x \to a}f(x)}}{{\displaystyle \lim_{x \to a}g(x) + A}}\textbf{ (correct)}$$ provided $\lim_{x \to a}g(x)$ exists and is not equal to $-A$.

Let us consider the following two examples: $$\lim_{x \to 0}\frac{x}{x + \log(1 + x)} = \lim_{x \to 0}\frac{x}{0 + \log(1 + x)} = 1\textbf{ (wrong)}$$ $$\lim_{x \to 0}\frac{x}{x^{2} + \log(1 + x)} = \lim_{x \to 0}\frac{x}{0 + \log(1 + x)} = 1\textbf{ (wrong)}$$ Both the examples use faulty manipulation as they replace $x$ and $x^{2}$ by the limit $0$ without splitting the limit operation till smaller parts. Out of these two examples first one has the wrong answer whereas (surprise!!) the second one has correct answer. People who are so fond of such replacements say that the replacement of $x^{2}$ by $0$ is valid because $x^{2}$ is of smaller order compared to $\log(1 + x)$ and hence can be safely put as $0$ whereas the same can't be done for $x$ (in first example) because it is of same order as that of $\log(1 + x)$.

I prefer to avoid such vague talk of orders and follow the right approach to both examples:
$$\lim_{x \to 0}\frac{x}{x + \log(1 + x)} = \lim_{x \to 0}\dfrac{1}{1 + \dfrac{\log(1 + x)}{x}} = \frac{1}{1 + 1} = \frac{1}{2}\textbf{ (correct)}$$ $$\lim_{x \to 0}\frac{x}{x^{2} + \log(1 + x)} = \lim_{x \to 0}\dfrac{1}{x + \dfrac{\log(1 + x)}{x}} = \frac{1}{0 + 1} = 1\textbf{ (correct)}$$ In both the examples the direct split of limits for division operation is not possible as the limit of denominator is zero and hence an algebraic manipulation is needed (division of both numerator and denominator by $x$) and then the limit operation is split across all parts (not shown above) and each part replaced by their limit to get the right answer.

However there are two occasions where it is possible to substitute an expression by its limit without affecting the final result:

1) If $h(x) = f(x) \pm g(x)$ and $\lim_{x \to a}g(x) = L$ exists then the behavior of $h(x)$ as $x \to a$ is same as the behavior of $f(x)$ as $x \to a$.

This means that if $f(x)$ tends to a limit (or diverges to $\pm\infty$, or oscillates finitely or infinitely) as $x \to a$, then so does $h(x)$ and hence while trying to find the limit of $h(x)$ we can substitute $g(x)$ by its limit $L$ without any problem.

2) If $h(x) = f(x) \cdot g(x)$ or $h(x) = f(x)/g(x)$ and $\lim_{x \to a}g(x) = L \neq 0$ exists then the behavior of $h(x)$ as $x \to a$ is same as the behavior of $f(x)$ as $x \to a$.

Note that the condition $L \neq 0$ is necessary here. Hence if $h(x) = f(x)\cdot g(x)$ or $h(x) = f(x)/g(x)$ and $\lim_{x \to a}g(x) = L \neq 0$ then we can replace $g(x)$ by its limit $L$ without affecting the calculation for limit of $h(x)$ as $x \to a$.

In the next post we will study certain advanced techniques for solving limit problems.