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First time posting here. I would like to get some help with this limit. I'm expected to solve it without using L'Hopital's rule, as I haven't been taught said rule, but I'm not sure how to go about it.

$$\lim_{x\to3} \frac{(\sin(x-3))}{(x^2-9)}$$

Could someone nudge me in the right direction? Thanks!

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  • $\begingroup$ Welcome. Hint: $\sin(x)$ is differentiable at $x=0$. $\endgroup$ Feb 23 at 13:36
  • $\begingroup$ Remember $\frac{\sin x}{x}, x\to 0$. $\endgroup$
    – zkutch
    Feb 23 at 13:36
  • $\begingroup$ Try factorizing the denominator and using $\lim_{x\to 0} \frac{\sin{x}}{x}=1$ $\endgroup$
    – Souparna
    Feb 23 at 13:37

2 Answers 2

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We have $$ \frac{\sin (x-3)}{x^2-9}=\frac{1}{x+3}\cdot\frac{\sin (x-3)}{x-3} $$ and $$ \lim_{x\to 3}\frac{1}{x+3}=\frac{1}{6}, $$ while $$ \lim_{x\to 3}\frac{\sin (x-3)}{x-3} =\lim_{t\to 0}\frac{\sin t}{t}=\cdots $$

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You can also use substitution $x-3=a \ ,\ a\to0 $ $$\lim_{x\to3} \frac{(\sin(x-3))}{(x^2-9)}\\x=a+3\\\lim_{a\to 0 }\frac {\sin a}{(a+3)^2-9}=\\\lim_{a\to 0 }\frac {\sin a}{a^2+6a+9-9}=\\\lim_{a\to 0 }\frac {\sin a}{a(a+6)}=?\\$$

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