Problem: find the limit
$lim _{x\rightarrow 0}\dfrac{\tan(3x)}{\sin(2x)}$
$lim _{x\rightarrow 0}\dfrac{(\sin(2x) + 3)}{(\cos(7x)-8)}$
Note
I am able to solve the first one using l'Hopitals, but I really want to be able to solve problems like this without it. Any tips/solutions much appreciated!
$\endgroup$1 Answer
$\begingroup$You should know the limits $\displaystyle\lim_{x \to 0}\dfrac{\tan x}{x} = \lim_{x \to 0}\dfrac{\sin x}{x} = 1$.
So, let's manipulate the first limit as follows:
$\displaystyle\lim_{x \to 0}\dfrac{\tan 3x}{\sin 2x} = \lim_{x \to 0}\dfrac{\frac{\tan 3x}{x}}{\frac{\sin 2x}{x}} = \lim_{x \to 0}\dfrac{3}{2}\dfrac{\frac{\tan 3x}{3x}}{\frac{\sin 2x}{2x}}$.
Do you see how to proceed from here?
For the second one, both the numerator and denominator are continuous for all $x$ and the denominator isn't zero. What does this tell you about the limit?
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