5
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Judging by this, I'd say your tutor is wrong. Neither of the limits will turn out to be $2^0$. $\endgroup$ – Hirshy Jul 22 '15 at 9:43
  • $\begingroup$ Good you got suspicious, since the calculation is just wrong. $\endgroup$ – Mr. Barrrington Jul 22 '15 at 9:43
1
$\begingroup$

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 $x=2$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ In fact, $f(2) = \mathrm{LHL} = 0$. However, there is still a discontinuity. $\endgroup$ – molarmass Jul 22 '15 at 10:13
  • $\begingroup$ yes, you are right $\endgroup$ – Harish Chandra Rajpoot Jul 22 '15 at 10:14
4
$\begingroup$

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 $x=2$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.