# Tutor profile: Sayantan M.

## Questions

### Subject: Linear Algebra

Is $$\{(x, y): x^{2}+y^{2} \leq 1\}$$ a subspace of $$\mathbb{R}^{2}$$?

let $$S = \{(x, y): x^{2}+y^{2} \leq 1\}$$. We see that $$(\frac{1}{2}, \frac{1}{2}) \in S$$ but $$4(\frac{1}{2}, \frac{1}{2}) = (2, 2) \notin S$$. Hence we conclude that S is not a subspce of $$\mathbb{R}^{2}$$.

### Subject: Basic Math

Is the function $$f(x) = |x|$$ differentiable at $$x=0$$?

Let us check the differentiability of the given function $$f(x) = \begin{cases} x & x>0 \\ -x & x<0 \\ 0 & x=0 \end{cases}$$ at $$x=0$$. We find out the left-hand derivative as well as the right-hand derivative of the given function at $$x=0$$.$$\text{L.H.D.} = \lim_{h\to 0-} \frac{f(0+h)-f(0)}{h} = \lim_{h\to 0-} \frac{-h}{h} = -1$$. $$\text{R.H.D.} = \lim_{h\to 0+} \frac{f(0+h)-f(0)}{h} = \lim_{h\to 0-} \frac{h}{h} = 1$$. We see that $$\text{L.H.D.} \neq \text{R.H.D.} $$. Hence we conclude that the function is not differentiable at $$x=0$$.

### Subject: Calculus

Does $$\lim_{x \to 0} x\sin (\frac{1}{x})$$ exist?

We know that $$-1 \leq \sin(\frac{1}{x}) \leq 1$$ which implies $$-x \leq x\sin(\frac{1}{x}) \leq x$$. Now taking limit as $$x\to 0 $$, we see that $$\lim_{x\to 0}(-x) = 0 = \lim_{x\to 0}(x)$$. Hence by the well known Sandwich theorem for limits, we conclude that $$\lim_{x \to 0} x\sin (\frac{1}{x}) =0$$ and hence the limit exists.

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