# continuity of a function

I have a task as preparation for my Calculus Exam.

$f(x)= \begin{cases} 2^{\frac{1}{x-2}} ,& x\neq 2 \\ 0 ,&x=2 \end{cases}$

Now we have the following solution by one of our tutors:

$l_1 = \lim_{x \rightarrow 2^-} f(x) = \lim_{x\rightarrow 2^-}2^{\frac{1}{x-2}} = 2^0 = 1$

$l_2 = \lim_{x \rightarrow 2^+} f(x) = \lim_{x\rightarrow 2^+}2^{\frac{1}{x-2}} = 2^0 = 1$

But I don't understand this specific part: $\lim_{x\rightarrow 2^+}2^{\frac{1}{x-2}} = 2^0 = 1$

What is she doing between that steps because if I have a fraction with $\dfrac{1}{\text{number} < 0}$ it is not getting $0$ but larger.

So where's the $2^0$ coming from?

• Judging by this, I'd say your tutor is wrong. Neither of the limits will turn out to be $2^0$. – Hirshy Jul 22 '15 at 9:43
• Good you got suspicious, since the calculation is just wrong. – Mr. Barrrington Jul 22 '15 at 9:43

Let's check continuity of the function $f(x)$ at $x=2$

Notice, $$LHL=\lim_{x\to 2^{-}}2^{\frac{1}{x-2}}$$ setting $x=2-h\implies h\to 0 \ as\ x\to 2$ $$LHL=\lim_{h\to 0}2^{\frac{1}{(2-h)-2}}$$ $$=\lim_{h\to 0}2^{\frac{-1}{h}}$$ $$=2^{(-\infty)}=0$$ Again notice $$RHL=\lim_{x\to 2^{+}}2^{\frac{1}{x-2}}$$ setting $x=2+h\implies h\to 0 \ as\ x\to 2$ $$RHL=\lim_{h\to 0}2^{\frac{1}{(2+h)-2}}$$ $$=\lim_{h\to 0}2^{\frac{1}{h}}$$ $$=2^{(\infty)}=\infty$$ & $$f(2)=0$$ $$\implies \color{blue}{f(2)=LHL\neq RHL}$$ Hence, the function $f(x)$ has discontinuity at $x=2$

• In fact, $f(2) = \mathrm{LHL} = 0$. However, there is still a discontinuity. – molarmass Jul 22 '15 at 10:13
• yes, you are right – Harish Chandra Rajpoot Jul 22 '15 at 10:14

Both $l_1$ and $l_2$ are wrong.

$$\begin{cases} l_1=\lim\limits_{x\to2^-} 2^\frac{1}{x-2}=2^{-\infty}=0\\ l_2=\lim\limits_{x\to2^+} 2^\frac{1}{x-2}=2^{+\infty}=+\infty\\ \end{cases}$$

Because $2^--2=0^-$ and $2^+-2=0^+$

$$l_1\not=l_2$$

Thus function is not continuous at $x=2$.

• As this is in preparation for a calculus exam, I wouldn't use the expression $2^\infty$, as this should be marked as incorrect. – Hirshy Jul 22 '15 at 9:46