# If \[\underset{x\to a}{\mathop{Lt}}\,\left[ f(x)g(x) \right]\] exists then both \[\underset{x\to a}{\mathop{Lt}}\,f(x)\]and $\underset{x\to a}{\mathop{Lt}}\,g(x)$ exist. State if True / False.

Last updated date: 19th Mar 2023

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Hint: Try remembering all the properties of limits, including the conditions that are imposed on them, to find out if this statement is true or false.

Let’s revisit all the properties of limit existence that we are aware of. However, we’ll revisit only the ones that involve some relation between two independent functions, having the same limit variable as their arguments.

So, let’s assume two functions first, that have $x$ as their argument, to make the functions look similar to the ones mentioned in the question. So, the two functions assumed can very well be $f(x)$ and $g(x).$

Now, let’s revisit all the properties of limits we know by now.

The first property says that $\underset{x\to a}{\mathop{\lim }}\,(f(x)+g(x))=\underset{x\to a}{\mathop{\lim }}\,f(x)+\underset{x\to a}{\mathop{\lim }}\,g(x)$.

The second property says that $\underset{x\to a}{\mathop{\lim }}\,(f(x)-g(x))=\underset{x\to a}{\mathop{\lim }}\,f(x)-\underset{x\to a}{\mathop{\lim }}\,g(x)$.

The third one says that $\underset{x\to a}{\mathop{\lim }}\,(f(x).g(x))=\underset{x\to a}{\mathop{\lim }}\,f(x).\underset{x\to a}{\mathop{\lim }}\,g(x)$.

The fourth one says that $\underset{x\to a}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}=\dfrac{\underset{x\to a}{\mathop{\lim }}\,f(x)}{\underset{x\to a}{\mathop{\lim }}\,g(x)}$.

And the last one says that $\underset{x\to a}{\mathop{\lim }}\,{{(f(x))}^{g(x)}}=\underset{x\to a}{\mathop{\lim }}\,{{(f(x))}^{\underset{x\to a}{\mathop{\lim }}\,g(x)}}$.

However, you should very carefully note one thing. For any of these properties to exist, it is mandatory for both, $\underset{x\to a}{\mathop{\lim }}\,f(x)$ and $\underset{x\to a}{\mathop{\lim }}\,g(x)$ to exist individually.

Now, let’s see what we have been given in the question.

In the question, we are asked for the existence of the individual limits, if their product exists, at the same limiting value.

In other words, we are told that $\underset{x\to a}{\mathop{\lim }}\,(f(x).g(x))$ exists.

Now, from the third property, we can say that, $\underset{x\to a}{\mathop{\lim }}\,(f(x).g(x))$ will exist if and only if both, $\underset{x\to a}{\mathop{\lim }}\,f(x)$ and $\underset{x\to a}{\mathop{\lim }}\,g(x)$ exist individually.

Thus, from here, we can conclude that since we’re given that $\underset{x\to a}{\mathop{\lim }}\,(f(x).g(x))$ exists, it automatically means that $\underset{x\to a}{\mathop{\lim }}\,f(x)$ and $\underset{x\to a}{\mathop{\lim }}\,g(x)$ exist individually too.

Hence, the Statement given to us in the question is True.

Note: Here, in these types of questions the mistake can happen interpreting which side of the property is the condition and which is the implication. For example, in these properties, the condition is that $\underset{x\to a}{\mathop{\lim }}\,f(x)$ and $\underset{x\to a}{\mathop{\lim }}\,g(x)$ exist individually, and the implications are the five properties stated above.

Let’s revisit all the properties of limit existence that we are aware of. However, we’ll revisit only the ones that involve some relation between two independent functions, having the same limit variable as their arguments.

So, let’s assume two functions first, that have $x$ as their argument, to make the functions look similar to the ones mentioned in the question. So, the two functions assumed can very well be $f(x)$ and $g(x).$

Now, let’s revisit all the properties of limits we know by now.

The first property says that $\underset{x\to a}{\mathop{\lim }}\,(f(x)+g(x))=\underset{x\to a}{\mathop{\lim }}\,f(x)+\underset{x\to a}{\mathop{\lim }}\,g(x)$.

The second property says that $\underset{x\to a}{\mathop{\lim }}\,(f(x)-g(x))=\underset{x\to a}{\mathop{\lim }}\,f(x)-\underset{x\to a}{\mathop{\lim }}\,g(x)$.

The third one says that $\underset{x\to a}{\mathop{\lim }}\,(f(x).g(x))=\underset{x\to a}{\mathop{\lim }}\,f(x).\underset{x\to a}{\mathop{\lim }}\,g(x)$.

The fourth one says that $\underset{x\to a}{\mathop{\lim }}\,\dfrac{f(x)}{g(x)}=\dfrac{\underset{x\to a}{\mathop{\lim }}\,f(x)}{\underset{x\to a}{\mathop{\lim }}\,g(x)}$.

And the last one says that $\underset{x\to a}{\mathop{\lim }}\,{{(f(x))}^{g(x)}}=\underset{x\to a}{\mathop{\lim }}\,{{(f(x))}^{\underset{x\to a}{\mathop{\lim }}\,g(x)}}$.

However, you should very carefully note one thing. For any of these properties to exist, it is mandatory for both, $\underset{x\to a}{\mathop{\lim }}\,f(x)$ and $\underset{x\to a}{\mathop{\lim }}\,g(x)$ to exist individually.

Now, let’s see what we have been given in the question.

In the question, we are asked for the existence of the individual limits, if their product exists, at the same limiting value.

In other words, we are told that $\underset{x\to a}{\mathop{\lim }}\,(f(x).g(x))$ exists.

Now, from the third property, we can say that, $\underset{x\to a}{\mathop{\lim }}\,(f(x).g(x))$ will exist if and only if both, $\underset{x\to a}{\mathop{\lim }}\,f(x)$ and $\underset{x\to a}{\mathop{\lim }}\,g(x)$ exist individually.

Thus, from here, we can conclude that since we’re given that $\underset{x\to a}{\mathop{\lim }}\,(f(x).g(x))$ exists, it automatically means that $\underset{x\to a}{\mathop{\lim }}\,f(x)$ and $\underset{x\to a}{\mathop{\lim }}\,g(x)$ exist individually too.

Hence, the Statement given to us in the question is True.

Note: Here, in these types of questions the mistake can happen interpreting which side of the property is the condition and which is the implication. For example, in these properties, the condition is that $\underset{x\to a}{\mathop{\lim }}\,f(x)$ and $\underset{x\to a}{\mathop{\lim }}\,g(x)$ exist individually, and the implications are the five properties stated above.

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