# Limits with sine and square roots

## Homework Statement

solve the limit: $$\lim_{x\rightarrow1}\frac{sin(x-1)}{\sqrt{x}-1}$$

## The Attempt at a Solution

Is there a way of how i can solve this without using l'hospital's rule or taylor series?

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Ray Vickson
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## Homework Statement

solve the limit: $$\lim_{x\rightarrow1}\frac{sin(x-1)}{\sqrt{x}-1}$$

## The Attempt at a Solution

Is there a way of how i can solve this without using l'hospital's rule or taylor series?
What is wrong with using l'Hospital's rule or Taylor series? Of course, changing to the variable $t = \sqrt{x}$ simplifies the question a lot.

RGV

Last edited:
Well i haven't learned l'hospitals rule or taylor's series so i can't use either of them.

Multiply numerator and denominator by $\sqrt{x}+1$.

Multiply numerator and denominator by $\sqrt{x}+1$.
ok i multiplied by √x+1 and i got:

$$\lim_{x\rightarrow1}\frac{sin(x-1)(\sqrt{x}+1)}{x-1}$$

now should i cancel the x-1?

Do you know the limit

$$\lim_{x\rightarrow 0}\frac{\sin(x)}{x}$$

Do you know the limit

$$\lim_{x\rightarrow 0}\frac{\sin(x)}{x}$$
yes it's 1

Now, use that to calculate the limit.

alright got it now:

$$\lim_{x\rightarrow1}\frac{sin(x)}{x}*(\sqrt{x}+1)=2$$

Is that correct?

Mark44
Mentor
alright got it now:

$$\lim_{x\rightarrow1}\frac{sin(x)}{x}*(\sqrt{x}+1)=2$$

Is that correct?
No.
$$\lim_{x\to 1}\frac{sin(x)}{x}=\frac{sin(1)}{1} \neq 1$$

Split your limit into two limits, and then do some fiddling with the first one so that you can use this limit:
$$\lim_{y \to 0}\frac{sin(y)}{y} = 1$$

No.
$$\lim_{x\to 1}\frac{sin(x)}{x}=\frac{sin(1)}{1} \neq 1$$

Split your limit into two limits, and then do some fiddling with the first one so that you can use this limit:
$$\lim_{y \to 0}\frac{sin(y)}{y} = 1$$
We still haven't learned how to split a limit into 2 different limits yet.

Mark44
Mentor
Are you sure? It's a very basic property of limits, and one that is presented pretty early.
$$\lim_{x \to a} f(x)\cdot g(x) = \lim_{x \to a}f(x) \cdot \lim_{x \to a}g(x)$$

The above is true as long as both limits on the right exist. As I see it, you pretty much need to use this idea in your problem.

Are you sure? It's a very basic property of limits, and one that is presented pretty early.
$$\lim_{x \to a} f(x)\cdot g(x) = \lim_{x \to a}f(x) \cdot \lim_{x \to a}g(x)$$

The above is true as long as both limits on the right exist. As I see it, you pretty much need to use this idea in your problem.
Yes I know that, but we haven't reached it yet in the lesson hence i can't use it.

Yes I know that, but we haven't reached it yet in the lesson hence i can't use it.
Then I'm afraid you can't solve this question.

You can't even solve questions like $\lim_{x\rightarrow 1}{x^2}$ in this case.

SammyS
Staff Emeritus
Homework Helper
Gold Member
ok i multiplied by √x+1 and i got:

$$\lim_{x\rightarrow1}\frac{sin(x-1)(\sqrt{x}+1)}{x-1}$$

now should i cancel the x-1?
Do you know the limit

$$\lim_{x\rightarrow 0}\frac{\sin(x)}{x}$$
yes it's 1
So, if $\displaystyle \lim_{x\rightarrow 0}\frac{\sin(x)}{x}=1\,,$ then what is $\displaystyle \lim_{x-1\rightarrow 0}\frac{\sin(x-1)}{x-1}\,?$