Question: Try evaluate below limit:
\lim_{x \to 0} (\frac{2^{x^{2}}+3^{x^{2}}}{2^x + 3^x})^{\frac{1}{x}}
Hint:
\lim_{x \to 0} u(x)^{v(x)} = e^{\lim_{x \to 0} [u(x)-1]v(x)}
Solution:
Taking the logarithm of the limit, we have:
\lim_{x \to 0} (\frac{2^{x^{2}}+3^{x^{2}}}{2^x + 3^x}-1)\frac{1}{x}
= \lim_{x \to 0} (\frac{2^{x^{2}}+3^{x^{2}} - 2^x - 3^x}{2^x + 3^x})\frac{1}{x}
= \lim_{x \to 0} (\frac{2^{x^{2}}+3^{x^{2}} - 2^x - 3^x}{x}) ‧ \lim_{x \to 0}\frac{1}{2^x+3^x}
= \lim_{x \to 0} (\frac{2^{x^{2}}+3^{x^{2}} - 2^x - 3^x}{x}) ‧ \frac{1}{2}
=\frac{1}{2} \lim_{x \to 0} (2x‧ln2‧2^{x^{2}} + 2x‧ln3‧3^{x^{2}} - ln2‧2^x - ln3‧3^x)
(by L'Hospital's Rule)
= ln\frac{1}{\sqrt{6}}
The desired limit is then \frac{1}{\sqrt{6}}.
Another Question:
Try evaluate \lim_{x \to 0} \frac{1-cosxcos2x...cosnx}{x^2}
Hint:
Rewrite the expression in exponent form:
\lim_{x \to 0} \frac{1-e^{lncosx + lncos2x + ... + lncosnx}}{x^2}
Solution:
By the hint, we have, using L'Hospital's Rule:
\lim_{x \to 0} \frac{- cosxcos2x...cosnx (-\frac{sinx}{cosx}-\frac{2sin2x}{cos2x}-....-\frac{nsinnx}{cosnx})}{2x}
using L'Hospital's Rule again, we have:
\frac{1}{2} \lim_{x \to 0} (\frac{sinx}{x}\frac{1}{cosx}+\frac{2^{2}sin2x}{x}\frac{1}{cos2x}+...+\frac{n^{2}sinnx}{x}\frac{1}{cosnx})
=\frac{1}{2} (1^{2}+2^{2} + ... + n^{2}) = \frac{n(n+1)(2n+1)}{12}
Before we go on, please read the following passages thoroughly:
e.g.
xlnx = o(x)
e^x = 1 + x + O(x^2)
... and some important freshman stuff as well:
Now you should be able to read it, and if not, Google your doubts in prompts.
Above is the proof.
Question:
Consider f(x) = x sinx
(a) Show that f(x) is an even function.
(b) Hence, show that f(x) has no upper bound and lower bound for all real x.
Solution:
(a) Both x and sinx are odd functions. Adding up their degrees will yield an even function.
Recall sin x = x + o(x), then x sin x = x^2 + o(x^2)
Q,E.D.
(b) y = x is always unbounded. Also, y = sinx is always oscillating. Their product will diverge from negative infinity to positive infinity.
Q.E.D.
Solution:
Higher math is all about gimmicks.
Notice the definition of a derivative:
f'(x_0)=\lim\limits\_{\Delta x\rightarrow 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}.
上面條題目係典型考古題嚟, 今次簡單啲:
Write the power series for \frac{1}{(1+x)^2}
Solution:
Remember this:
e^{-\infty} = 0
So we have:
Left hand limit= Right hand limit
The required limit is then 1.
This time we will use Mean Value Theorem to compute a limit. Try this:
