Question

1) Evaluate lim x->1 (x-1/Sqrt(x)-1)
2)Evaluate limx->0 (1-e^(2x)/1-e^x)

Answer 1

\[lim_{x \to 1} \frac{x-1}{\sqrt{x}-1}\]For the first one?

Answer 2

yup

Answer 3

find the first derivative of numerator and denominator and put value of limit x -> 1 your will get your answer

Answer 4

Hmm.. multiply a conjugate\[ \lim_{x \to 1} \frac{x-1}{\sqrt{x}-1}\]\[=\lim_{x \to 1} \frac{x-1}{\sqrt{x}-1}\times \frac{\sqrt{x}+1}{\sqrt{x}+1}\]\[=\lim_{x \to 1} \frac{(x-1)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}\]Expand the bottom part.

Answer 5

I guess it tends to 2.


(x-1)=(sqrt{x}-1)(sqrt{x}+1}

Answer 6

Use \(a^2 - b^2 = (a+b)(a-b)\) to expand it. Cancel what you can cancel

Answer 7

You do not have to do the conjugate see this....\[\lim_{x \to 1} \frac{x-1}{\sqrt{x}-1}\]
\[\lim_{x \rightarrow 1}\frac{ (\sqrt{x}-1)(\sqrt{x}+1) }{ \sqrt{x}-1 }\]

Answer 8

cant you just use l hospitals rule on problem 1?

Answer 9

No

Answer 10

Now cancel the terms from the numerator and denominator @Tech-Hunter

Answer 11

Then apply limit and kaboom you get your answer.