# Thread: Limits Problems

1. ## Limits Problems

Hi. I need help. My test is next week, and on it there will be a lot of math stuff including limits. Now, the teacher gave us a few problems to show us what to expect on the test. That was last week. I've been trying to solve these limits:

but without success.

Here's the catch: I can't use l'Hopital's Rule.

I would really appreciate some solutions. Thx in advance.

P.S. You don't have to write the whole solutions, just the general idea.

2. For the "sin" problem, multiply numerator and denominator by $\sqrt{y}+1$ and note that $\lim_{y\to0}\frac{\sin y}{y}=1$, the answer will be $-2$

For the "ln" problem, note that $x$ approaches more rapidly to $0$ than $\ln x$, so the answer becomes $0$.

The other one--algebraic manipulation.

$\displaystyle \lim_{x\to4} \frac{x-4}{\sqrt[3]{x+23} - 3}$

First, recall the identity: . $a^3 - b^3 \:=\:(a-b)(a^2+ab + b^2)$

We have: . $\displaystyle f(x) \;=\;\frac{x-4}{(x+23)^{\frac{1}{3}} - 3}$

Multiply numerator and denominator by: $(x+23)^{\frac{2}{3}} + 3(x+23)^{\frac{1}{3}} + 9$

$\displaystyle f(x) \;=\;\frac{x-4}{(x+23)^{\frac{1}{3}} - 3}\cdot\frac{(x+23)^{\frac{2}{3}} + 3(x+23)^{\frac{1}{3}} + 9} {(x+23)^{\frac{2}{3}} + 3(x+23)^{\frac{1}{3}} + 9}$

. . . . $\displaystyle =\;\frac{(x-4)\left[(x+23)^{\frac{2}{3}} + 3(x+23)^{\frac{1}{3}} + 9\right]}{(x+23) - 3^3}$

. . . . $\displaystyle =\;\frac{(x-4)\left[(x+23)^{\frac{2}{3}} + 3(x+23)^{\frac{1}{3}} + 9\right]}{x-4}$

. . . . $\displaystyle =\;(x+23)^{\frac{2}{3}} + 3(x+23)^{\frac{1}{3}} + 9$

Therefore: . $\displaystyle \lim_{x\to4}\left[(x+23)^{\frac{2}{3}} + 3(x+23)^{\frac{1}{3}} + 9\right]$

. . . . . . . . . . $=\; 27^{\frac{2}{3}} + 3\cdot27^{\frac{1}{3}} + 9$

. . . . . . . . . . $=\;9 + 3\cdot3 + 9$

. . . . . . . . . . $= \;27$

4. Setting $x=1-\xi$ the limit becomes...

$\displaystyle \lim_{x \rightarrow 0} x\ \ln x = \lim_{\xi \rightarrow 1} (1-\xi)\ \ln (1-\xi)$ (1)

Now is...

$\displaystyle \ln (1-\xi) = -\xi - \frac{\xi^{2}}{2} - \frac{\xi^{3}}{3} - ...$ (2)

... so that...

$\displaystyle (1-\xi)\ \ln (1-\xi) = -\xi\ (1-\xi) - \frac{\xi^{2}}{2}\ (1-\xi) - \frac{\xi^{3}}{3}\ (1-\xi) - ...$ (3)

If $\xi = 1$ each term of the 'infinite sum' (3) is 0 so that the limit of 'infinite sum' is 0...

Kind regards

$\chi$ $\sigma$

5. You guys are amazing... I love this forum.

Thank you so much. Thanks allround!