I'm doing a few past exam papers for my calculus test. I came across a few problems which I thought would be worth asking about.

1.) Consider the function

$f(x) = \begin{cases} x+1, & \text{if $x$ < 1} \\ b, & \text{if $x$ = 1} \\ -(x-1)^2 & \text{if $x$ > 1} \end{cases}$

The limit $\lim\limits_{x \to 1}f(x)$ exists for ?

Would the correct approach be to evaluate the limit from the left and right , see that they are not equal and thus deduce that the limit as x approaches 1 for f(x) does not exist?

2.) $\lim\limits_{x \to 1} {sin(x -1 ) \over (x^2-1)}$

Direct substitution won't work because we get a zero as the denominator. If I remember correctly from class there are smaller proofs involving trigonometry and limits which we used to prove this ( I'll post what we did if I find it )

At first I considered using LHospitals rule but the expression isn't in the correct form so I entered it into wolfram alpha and got 1/2. I can get to the answer by evaluating the bottom part of the expression as x approaches 1 as such :

$Let f(x) = x^2 - 1$

$f(x) = (x^2 -1)$

$f'(x) = 2x$

$\lim\limits_{x \to 1}f'(x) = 2$

And then simply directly substituting for the upper part of the equation. Finally getting to $1 \over 2$ as the answer. This feels very off though and I doubt it's correct ( the method at least ).

3.) Let $y = x^x$

Find y'.

I didn't get far with this question , I was thinking about using taking the Ln on both sides and then doing implicit differentiation. Would that lead me to the answer or is there an easier way?

Thanks for all your time :)

  1. You are right with the first one.
  2. According to L'Hopital's Rule, you need to differentiate both the numerator and denominator.

$\lim_{x\to1}\frac{\sin(x-1)}{x^2-1}=\lim_{x\to1}\frac{\cos(x-1)}{2x}=\frac{\cos(1-1)}{2\times 1}=\frac{1}{2}$

  1. Take logarithm of both sides and them use properties of logarithm functions.

$\ln(y)=\ln(x^x)$$\Rightarrow \ln(y)=x\ln(x)$

Take derivative of both sides.

$y'\frac{1}{y}=\ln(x) + x\frac{1}{ x}=\ln(x)+1$

$\Rightarrow y'=(\ln(x)+1)y$

Substitute y with $x^x$


  1. That is the correct approach. The limit exists if and only if exist both one-sided limits, and they are equal.

  2. You can use L'Hospital by doing $f(x)=sin(x-1), g(x)=x^2-1$, and $f,g$ are differentiable, and $\lim_{x\rightarrow 1} f(x)=0=\lim_{x\rightarrow 1} g(x)$. Also, exists the limit $\lim_{x\rightarrow 1} \frac{f'(x)}{g'(x)}=\lim_{x\rightarrow 1}\frac{\cos(x-1)}{2x}=\frac{1}{2}$, and therefore, you can use L'Hospital.

  3. $y=x^x=e^{ln(x^x)}=e^{xln(x)}=e^xx$, such that $y'=e^x+xe^x$

  1. Indeed, the limits from the left and right disagree, so there is no limit
  2. L'Hôpital should work here
  3. $y = x^x = e^{\ln(x^x)} = e^{x \ln(x)}$ (for $x > 0$)


For question $3)$, use the fact that $$f(x)=x^x=e^{x\ln x}$$ and $$(e^{g(x)})'=g'(x)e^{g(x)}$$


There is no need to use L'hopital for the second question. Observe that $$\lim_{x\to 1}\frac{\sin(x-1)}{x^2-1}=\lim_{x\to 1}\frac{\sin(x-1)}{x-1}\lim_{x\to 1}\frac{1}{x+1}=1\times\frac{1}{2}=\frac{1}{2}$$

  • $\begingroup$ I was about to say this; Some calculus courses forbid usage of L'Hospital. $\endgroup$ – user305860 May 31 '16 at 17:42
  • $\begingroup$ I'm not really sure what you're doing would you kindly explain? :) Are you factoring $x^2 -1$ ? $\endgroup$ – Armand Connor Du Plooy Jun 1 '16 at 5:59

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