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Evaluate the following limit without L'Hopital's rule!

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Determine the following limit - but do not use L'Hopital's rule:

$$\lim_{x\rightarrow 9}\frac{sin(\sqrt{x}-3)}{x-9}$$  Hint: Let $$t =\sqrt{x}$$

I would really appreciate some assistance on this. I know that as x-> 9, t->3, but I don't know how to adjust my limit to suit this. Thanks

Sep 17, 2022

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Let's use the hint, $$\text{If} \space \space t=\sqrt{x} \iff t^2=x$$

We get: $$lim_{t->3}\dfrac{sin(t-3)}{t^2-9}$$

Factor the denominator:

$$lim_{t->3}\dfrac{sin(t-3)}{(t-3)(t+3)}$$

Then, let's rewrite this as:

$$lim_{t->3}[\dfrac{sin(t-3)}{(t-3)}*\dfrac{1}{t+3}]$$

Using the following law of limits: $$lim_{x->a}[g(x)f(x)]=(lim_{x->a}g(x))*(lim_{x->a}f(x))$$

So we get:

$$lim_{t->3}[\dfrac{sin(t-3)}{(t-3)}*\dfrac{1}{t+3}]=[lim_{t->3}\dfrac{sin(t-3)}{(t-3)}]*[lim_{t->3}\dfrac{1}{t+3}]$$

Recall: $$lim_{y->0}\frac{sin(y)}{y}=1$$  (You could find the proof of this on youtube, but this is usually a standard result; that is, when you are solving a question, you do not need to prove this.)

Now, (t-3) as t-->3 will go to 0 and (t-3) as t goes to 3 will go to 0 as well, so basically this is the same as: $$lim_{y->0}\frac{sin(y)}{y}=1$$

(Where y=t-3). Hence, the limit in the first bracket is just 1, and the second limit in the other bracket could be done by direct substitution.

So:

$$1*(\dfrac{1}{3+3})=\dfrac{1}{6}$$ which is our answer! :)
I hope this helps!

Sep 17, 2022