How do you find #lim (sqrtx-1)/(x-1)# as #x-1^+# using l'Hospital's Rule or otherwise?

Question

How do you find #lim (sqrtx-1)/(x-1)# as #x->1^+# using l'Hospital's Rule or otherwise?

Samantha Adkison · Accepted Answer

I wouldn't use l'Hospital for this.

I would factor: #x-1 = (sqrtx+1)(sqrtx -1)# #lim_(xrarr1^+) (sqrtx-1)/(x-1) = lim_(xrarr1) (sqrtx-1)/((sqrtx-1)sqrtx+1)# # = lim_(xrarr1) 1/(sqrtx+1) = 1/(1+1) = 1/2# (Of course, I wouldn't write all that if I wasn't trying to explain. #lim_(xrarr1^+)(sqrtx-1)/(x-1) = lim_(xrarr1)1/(sqrtx+1) = 1/2# OR "Rationalize the numerator by multiplying by #(sqrtx+1)/(sqrtx+1)# to get #lim_(xrarr1^+) (sqrtx-1)/(x-1) = lim_(xrarr1) overbrace(cancel(x-1))^1/(cancel((x-1)) (sqrtx+1)) = 1/2#

Evelyn Agard · Answer

undefined To find $\lim_{x \to 1^+} \frac{\sqrt{x} - 1}{x - 1}$, you can apply L'Hôpital's Rule or rewrite the expression to eliminate the indeterminate form.

Using L'Hôpital's Rule, differentiate the numerator and the denominator separately:

$f(x) = \sqrt{x} - 1$

$g(x) = x - 1$

Then apply L'Hôpital's Rule:

$\lim_{x \to 1^+} \frac{f(x)}{g(x)} = \lim_{x \to 1^+} \frac{f'(x)}{g'(x)}$

$\lim_{x \to 1^+} \frac{\frac{1}{2\sqrt{x}}}{1} = \frac{1}{2}$

Alternatively, you can rewrite the expression:

$\frac{\sqrt{x} - 1}{x - 1} = \frac{(\sqrt{x} - 1)(\sqrt{x} + 1)}{(x - 1)(\sqrt{x} + 1)} = \frac{x - 1}{(x - 1)(\sqrt{x} + 1)} = \frac{1}{\sqrt{x} + 1}$

Taking the limit as $x$ approaches $1$ from the positive side:

$\lim_{x \to 1^+} \frac{1}{\sqrt{x} + 1} = \frac{1}{\sqrt{1} + 1} = \frac{1}{2}$

How do you find #lim (sqrtx-1)/(x-1)# as #x-&gt;1^+# using l'Hospital's Rule or otherwise?

How do you find #lim (sqrtx-1)/(x-1)# as #x->1^+# using l'Hospital's Rule or otherwise?