Limit - Practice Questions & MCQ

Limit

Consider the function $f(x)=x^2$

Observe that as $x$ takes values very close to 0, the value of $f(x)$ is also close to 0. (See graph below)
We can also interpret it in another way. If we input the values of $x$ which tend to/approach 0 (meaning close to 0, either just smaller than 0 or just larger than 0 ), the value of $f(x)$ will tend to 0 /approach (meaning close to 0, either just smaller than 0 or just larger than 0 ). Note we do NOT want to see what happens at $x=0$, we just want values of $x$ which are very close to 0.

Then we can say that, $\underset{x\to 0}{lim}f(x)=0$. (to be read as limit of $f(x)$ as $x$ tends to zero equals zero).
Similarly, when $x$ approaches 2 , the value of $f(x)$ approaches 4 , i.e. $\underset{x\to 2}{lim}f(x)=4$ or $\underset{x\to 2}{lim}{x}^2=4$

In General, as $x\to a$ (read as $x$ tends to $a),f(x)\to l$, then then l is called limit of the function $\mathrm{f}(\mathrm{x})$ which is symbolically written as $\underset{x\to a}{lim}f(x)=l$

Now consider the function.

$f(x)=\frac{x^2-6x-7}{x-7}$
We can factor the function as shown.

$\begin{array}{rl}& \mathrm{f}(\mathrm{x})=\frac{(\mathrm{x}-7)(\mathrm{x}+1)}{\mathrm{x}-7}\\ & \mathrm{f}(\mathrm{x})=\mathrm{x}+1,\mathrm{x}\ne 7\end{array}$

[Cancel like factors in numerator and denominator.] graphically

What happens at $x=7$ is completely different from what happens at points close to $x=7$ on either side. Just observe that as the input $x$ approaches 7 from either the left or the right, the output approaches 8. The output can get as close to 8 as we like if the input is sufficiently near 7. So we say that limit of $\underset{x\to 7}{lim}f(x)=8$ this function at $x=7$ equals 8 and it is denoted by $x\to 7$

So even if the function does not exist at $x=a$, still the limit can exist at that point as the limit is concerned only about the points close to $x=a$ and NOT at $x=a$ itself.