Answer
(a)
As $h$ tends to $0$, both $g(h)$ and $g(ah)$ tends to same value of function $g$.
Hence, $\lim\limits_{h \to 0}g(ah)=L$ for any constant $a\ne0$.
(b)
$g(1)$ and $g(a)$ can either be equal or not equal.
Hence, it is not necessarily true that $\lim\limits_{h \to 1}g(ah)=L$.
(c)
$\lim\limits_{h \to 0}f(h)=f(0)=0$
Also, $\lim\limits_{h \to 0}f(ah)=f(0)=0$.
$\lim\limits_{h \to 1}f(h)=f(1)=1$ but $\lim\limits_{h \to 1}f(ah)=f(a)$.
It is not necessarily true that $\lim\limits_{h \to 1}f(ah)=1$.
Work Step by Step
(a)
If $h$ tends to $0$ then $g(h)$ tends to $g(0)$.
which can also be written as $\lim\limits_{h \to 0}g(h)=g(0)$.
As we know that $\lim\limits_{h \to 0}g(h)=L$.
This gives $g(0)=L$.
If $h$ tends to $0$ then $a\times h$ also tends to $0$ for any constant $a\ne0$.
Thus, if $h$ tends to $0$ then $g(ah)$ tends to $g(0)$.
which can be written as $\lim\limits_{h \to 0}g(ah)=g(0)$.
As $g(0)=L$, $\lim\limits_{h \to 0}g(ah)=L$.
As $h$ tends to $0$, both $g(h)$ and $g(ah)$ tends to same value of function $g$.
Hence, $\lim\limits_{h \to 0}g(ah)=L$ for any constant $a\ne0$.
(b)
If instead $\lim\limits_{h \to 1}g(h)=L$.
Then $g(1)=L$.
As $\lim\limits_{h \to 1}g(ah)=g(a)$ and $a$ can be any constant.
$g(1)$ and $g(a)$ can either be equal or not equal.
Hence, it is not necessarily true that $\lim\limits_{h \to 1}g(ah)=L$.
(c)
It is given that $f(x)=x^{2}$.
On substituting $x=h$ we get $f(h)=h^{2}$.
Now substitute $h=0$.
We get $f(0)=0^{2}=0$.
Now again substitute $h=1$.
We get $f(1)=1^{2}=1$
Now solve limits as follows:
$\lim\limits_{h \to 0}f(h)=f(0)=0$
Also, $\lim\limits_{h \to 0}f(ah)=f(0)=0$.
$\lim\limits_{h \to 1}f(h)=f(1)=1$ but $\lim\limits_{h \to 1}f(ah)=f(a)$.
As $a$ can be any constant, it is not necessarily true that $f(a)=1$.
Hence, it is not necessarily true that $\lim\limits_{h \to 1}f(ah)=1$.
