# Thread: I Need Help Applying L'Hospital's Rule

1. ## I Need Help Applying L'Hospital's Rule

Problem #1

lim .... ln(x)
x->0+ ----
.......... x

This appears to have a form of negative infinity over zero. I do not see how to simplify, or apply L'Hospital's rule.

Problem #2

lim ... x^a -a^x + a -1
x->1 ----------------
............ (x-1)^2

This has a form of 0/0, thus I apply L'Hospital's rule:

= lim ... ax^(a-1) - a^x * ln(a)
x->1 .. ------------------------
..................... 2(x-1)

and I simplify:

= lim .. ax^a - x* a^x * ln(a)
x->1 . ------------------------
.................... 2x(x-1)

The denominator goes to zero, but the numerator goes to a - a * ln(a). Again I am stuck.

Any help is appreciated.

If someone could give me a concise algorithm which would help me always find the limits(at this level of math), I would be grateful.

2. Originally Posted by masonbrewey

Problem #1

lim .... ln(x)
x->0+ ----
.......... x

This appears to have a form of negative infinity over zero. I do not see how to simplify, or apply L'Hospital's rule.

Problem #2

lim ... x^a -a^x + a -1
x->1 ----------------
............ (x-1)^2

This has a form of 0/0, thus I apply L'Hospital's rule:

= lim ... ax^(a-1) - a^x * ln(a)
x->1 .. ------------------------
..................... 2(x-1)

and I simplify:

= lim .. ax^a - x* a^x * ln(a)
x->1 . ------------------------
.................... 2x(x-1)

The denominator goes to zero, but the numerator goes to a - a * ln(a). Again I am stuck.

Any help is appreciated.

If someone could give me a concise algorithm which would help me always find the limits(at this level of math), I would be grateful.
in the first one let
$y=ln(x) \Rightarrow e^y=x$ so when $x\rightarrow 0^+$ $....\;\;\quad y\rightarrow -\infty$

$lim_{y\rightarrow -\infty}\frac{y}{e^y} = lim_{y\rightarrow -\infty} y(e^{-y})$

the second one

$lim_{x\rightarrow 1}\frac{x^a-a^x+a-1}{(x-1)^2}=\frac{0}{0}=lop=lim_{x\rightarrow 1}\frac{ax^{a-1}-a^x(lna)}{2(x-1)}=\frac{a-aln(a)}{0}=\infty$

3. Amer,

Thank you for your help. However, according to the answer sheet, the answer to the first is negative infinity and 1/2*a(a - 1) for the second.

4. Originally Posted by Amer
in the first one let
$y=ln(x) \Rightarrow e^y=x$ so when $x\rightarrow 0^+$ $....\;\;\quad y\rightarrow +\infty$

$lim_{y\rightarrow +\infty}\frac{y}{e^y} = \frac{+\infty}{+\infty} =lop=lim_{y\rightarrow +\infty}\frac{1}{e^y}=0$

the second one

$lim_{x\rightarrow 1}\frac{x^a-a^x+a-1}{(x-1)^2}=\frac{0}{0}=lop=lim_{x\rightarrow 1}\frac{ax^{a-1}-a^x(lna)}{2(x-1)}=\frac{a-aln(a)}{0}=\infty$
I believe you'll find that $\lim_{x \to 0^{+}}{\ln{x}} = -\infty$...

5. sorry ...