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, $\lim _{x \rightarrow 0} f(x)=0$. (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. $\lim _{x \rightarrow 2} f(x)=4 \quad$ or $\quad \lim _{x \rightarrow 2} x^2=4$

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

Now consider the function.

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

$
\begin{aligned}
& \mathrm{f}(\mathrm{x})=\frac{(\mathrm{x}-7)(\mathrm{x}+1)}{\mathrm{x}-7} \\
& \mathrm{f}(\mathrm{x})=\mathrm{x}+1, \mathrm{x} \neq 7
\end{aligned}
$

[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 $\lim _{x \rightarrow 7} f(x)=8$ this function at $x=7$ equals 8 and it is denoted by $x \rightarrow 7$

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.