Functions — Question 5

Question 5

A function $f$ is defined as: $$f(x) = \begin{cases} \dfrac{\ln(1 + ax)}{x}, & x \neq 0, \\[0.6em] 1, & x = 0. \end{cases}$$

(a) Find the value of $a$ for which $f(x)$ is continuous at $x = 0$.

(b) For that value of $a$, determine whether $f$ is differentiable at $x = 0$.

Question 5 - Solution

(a) For $f$ to be continuous at $x=0$, we need $$\lim_{x\to 0} f(x)=f(0)=1.$$ For $x\neq 0$, $$f(x)=\frac{\ln(1+ax)}{x}.$$ So compute $$\lim_{x\to 0}\frac{\ln(1+ax)}{x}.$$ As $x\to 0$, the numerator and denominator both approach $0$, so we use L’Hospital’s Rule: $$\lim_{x\to 0}\frac{\ln(1+ax)}{x} = \lim_{x\to 0}\frac{\frac{a}{1+ax}}{1} = a.$$ For continuity, this limit must equal $f(0)=1$. Therefore, $$a=1.$$ Thus, $$\boxed{a=1}.$$

(b) Now let $a=1$. Then $$f(x)=\frac{\ln(1+x)}{x}, \qquad x\neq 0,$$ and $f(0)=1$.

To check differentiability at $x=0$, use the definition: $$f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h}.$$ Substitute $f(h)=\dfrac{\ln(1+h)}{h}$ and $f(0)=1$: $$f'(0) = \lim_{h\to 0} \frac{\frac{\ln(1+h)}{h}-1}{h} = \lim_{h\to 0} \frac{\ln(1+h)-h}{h^2}.$$ This is also a $0/0$ form, so apply L’Hospital’s Rule: $$f'(0) = \lim_{h\to 0} \frac{\frac{1}{1+h}-1}{2h} = \lim_{h\to 0} \frac{-h}{2h(1+h)} = \lim_{h\to 0} \frac{-1}{2(1+h)} = -\frac{1}{2}.$$ Since this limit exists, $f$ is differentiable at $x=0$, and $$\boxed{f'(0)=-\frac{1}{2}}.$$