If the function f defined as f(x)=1x-k-1e^2x-1, x0, is continuous at x=0, then the ordered…

Question

If the function $f$ defined as $f(x)=\dfrac1x-\dfrac{k-1}{e^{2x}-1}$, $x
e0$, is continuous at $x=0$, then the ordered pair $(k,f(0))$ is equal to?

Accepted Answer

$$\lim_{x\to0}\frac{e^{2x}-1-kx+x}{x(e^{2x}-1)}=\lim_{x\to0}\frac{(3-k)x+2x^2+\frac43x^3+\cdots}{2x^2+2x^3+\cdots}$$
For the limit to exist, need $3-k=0\Rightarrow k=3$. Then the limit reduces to $\dfrac{2}{2}=1=f(0)$.

So $(k,f(0))=(3,1)$.

Solutions · 1