# Limit as x approaches some constant value

I have some problems with calculating limits as x approach some const value. I have to find the value of limits $$\lim_{x \to -2} \frac{x^2+x-2}{x^2-3x+2}$$ $$\lim_{x \to 0} \frac{\sin7x}{\sin13x}$$ I see that the first value should be $$0$$? Because as we take $$x$$ really close to $$-2$$, values are approaching $$0$$. And I think that the limit of second case, should be $$\frac{7}{13}$$? But i have no idea how to prove it formally, using the delta, epsilon definition (Without L'Hostpital rule) Could someone explain and show me how to do it formally, by definition?

• Use algebraic rewriting and known properties instead of epsilon and delta. For the second one, see math.stackexchange.com/questions/565286/… (and countless similar questions). – Hans Lundmark Jan 8 '20 at 13:47
• Are you sure you don't have $x^2+3x+2$ in denominator? – bjorn93 Jan 8 '20 at 13:48

For the first one, since the function is continuous at $$x=-2$$, the limit is equal to the evaluation of the function at $$-2$$.
$$\lim_{x \to -2} \dfrac{x^2+x-2}{x^2-3x+2} = \dfrac{(-2)^2+(-2)-2}{(-2)^2-3(-2)+2} = \dfrac{0}{12} = 0$$
\begin{align*}\lim_{x\to 0} \dfrac{\sin 7x}{\sin 13x} & = \lim_{x \to 0} \dfrac{\tfrac{7x}{7x}}{\tfrac{13x}{13x}}\cdot \dfrac{\sin 7x}{\sin 13x} \\ & = \lim_{x \to 0} \dfrac{7x}{13x}\cdot \dfrac{\tfrac{\sin 7x}{7x}}{\tfrac{\sin 13x}{13x}} \\ & = \left( \lim_{x \to 0} \dfrac{7x}{13x}\right)\cdot \dfrac\lim_{x \to 0} \dfrac{\sin 7x}{7x}}\lim_{x \to 0} \dfrac{\sin 13x}{13x}} \\ & = \dfrac{7}{13}\cdot \dfrac{1}{1} = \dfrac{7}{13}\end{align*}
Here I used $$\displaystyle \lim_{u \to 0} \dfrac{\sin u}{u} = 1$$, the product rule for limits (so long as the limits of both multiplicands exist, then the product of the limits is the limit of the product), and the quotient rule for limits (so long as the limits of both numerator and denominator exist and the limit of the denominator is nonzero, then the quotient of the limits is the limit of the quotient).