$$ \lim_{x\to 0}\frac{\sqrt{1+2x}-\sqrt[\Large3]{1+5x}}{\sqrt[\Large5]{1+x}-\sqrt[\Large5]{1+2x}}$$
Multiplication by conjugate hasn't worked. I need a hint to start.
Asked
Active
Viewed 80 times
2
-
1Are you familiar with Taylor series/approximations? – Clement C. Oct 16 '15 at 17:19
-
Suppose you write the first term in the numerator as $(1+2x)^{1/2}=\left((1+2x)^{3/2}\right)^{1/3}$. Then perhaps you could use the well-known identity,$$a^n-b^n=a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}$$ – user170231 Oct 16 '15 at 17:19
-
L'Hoptial might help. (It looks like it'd be very messy, though.) – Akiva Weinberger Oct 16 '15 at 17:22
-
@ClementC. it's forbidden for this task. And I'm not familiar. – Val Oct 16 '15 at 17:23
-
Ah, Then my answer becomes sort of useless to you; but what are you allowed, then? (stating in your question what you can't use may be relevant) – Clement C. Oct 16 '15 at 17:25
-
Another suggestion, courtesy of Did. – user170231 Oct 16 '15 at 17:28
-
@user170231: skimming it, it does rely on Taylor expansions (even though only up to first order), doesn't it? – Clement C. Oct 16 '15 at 17:29
-
@ClementC. Yes, my mistake. I was in the process of searching for that older question/linking it in a comment when I heard the news. – user170231 Oct 16 '15 at 17:31
-
@user170231 well, in any case it's a useful reference, to keep at hand. – Clement C. Oct 16 '15 at 17:32
3 Answers
3
Applying L'Hospital's rule, we get
$$\lim_{x\to 0}\frac{\sqrt{1+2x}-\sqrt[3]{1+5x}}{\sqrt[5]{1+x}-\sqrt[5]{1+2x}}$$
$$=\lim_{x\to 0}\frac{\frac{2}{2\sqrt{1+2x}}-\frac{5}{3\sqrt[2/3]{1+5x}}}{\frac{1}{5\sqrt[4/5]{1+x}}-\frac{2}{5\sqrt[4/5]{1+2x}}}$$
$$=\frac{\frac{2}{2}-\frac{5}{3}}{\frac{1}{5}-\frac{2}{5}}=\frac{\frac{-2}{3}}{\frac{-1}{5}}=\color{red}{\frac{10}{3}}$$
Harish Chandra Rajpoot
- 37,450
-
1
-
Rule of thumb: When your numerator and denominator both go to zero, L'Hospital's Rule is the first thing you should try. – Jerry Guern Oct 16 '15 at 17:30
-
@JerryGuern That's a matter of taste. My thumb is leaning towards Taylor. – Clement C. Oct 16 '15 at 17:31
-
@ClementC. Your Taylor method below takes a bit more know-how than a student might have. L'Hospital's Rule is harder to mess up. And anyway, L'Hospital's Rule just amounts to Taylor-expanding the numerator and denominator separately so you can cancel some x's before plugging in x=0. – Jerry Guern Oct 16 '15 at 17:38
-
@JerryGuern The first thing I think of is the definition of the derivative. L'Hopital is often just a black box that gives no insight. (And I disagree that L'Hopital amounts to Taylor.) – zhw. Oct 16 '15 at 18:23
-
@zhw. If L'Hopital is a black box to you and you gained no insight from it, I suggest studying it more closely. Try this exercise. To calculate the limit of f(x)/g(x) as x->0, Taylor-expand f and g separately. If f(0) and g(0) are both 0, you will be able to cancel x from the numerator and denominator. In fact if the 0-nth derivs of f and g all equaled 0 at x=0, you'll be able to cancel x^(n+1). The fraction that remain will clearly approach $f^{(n+1)}(0)/g^{(n+1)}(0)$. And that's L'Hopital's rule, derived from Taylor expansion. Make sense? – Jerry Guern Oct 16 '15 at 18:55
1
Using Taylor series: when $x\to0$, we have, for any fixed $\alpha$, $(1+x)^\alpha = 1+\alpha x + o(x)$.
In your case, $$ \frac{(1+2x)^{1/2} - (1+5x)^{1/3}}{(1+x)^{1/5}-(1+2x)^{1/5}} = \frac{1+x + o(x) - (1+\frac{5}{3}x + o(x)))}{1+\frac{1}{5}x + o(x) -(1+\frac{2}{5}x + o(x))} = (\cdots) $$
Edit: in a subsequent comment, it turns out the OP is not allowed to use Taylor series. I'm leaving this here, however, as it may be useful to other readers who don't have this constraint. (?)
Clement C.
- 67,323
-
-
@ValeryBaturin well, Taylor expansion to first order is basically a linear approximation: $f(x) \approx f(0) + f^\prime(0)x$. (And that's all that is needed here). – Clement C. Oct 16 '15 at 17:33
-
0
Just using the definition of the derivative: Let $a(x) = (1+2x)^{1/2},$ $ b(x) = (1+5x)^{1/3},$ $ c(x) = (1+x)^{1/5} ,d(x) = (1+2x)^{1/5}.$ The expression equals $$\frac{a(x) - b(x)}{c(x) - d(x)} = \frac{(a(x)-a(1)) - (b(x)-b(1))}{(c(x)-c(1)) - (d(x)-d(1))}$$ $$ = \frac{(a(x)-a(1))/(x-1) - (b(x)-b(1))/(x-1)}{(c(x)-c(1))/(x-1) - (d(x)-d(1))/(x-1)}.$$
By definition of the derivative, the above has limit $$\frac{a'(1)-b'(1)}{c'(1)-d'(1)},$$
an easy computation.
zhw.
- 105,693
-
1
-
@martycohen I disagree. You can't derive L'Hopital this way, because in LHR there is no assumption that derivatives exist at the point in question. Note also this problem could be solved in the first few weeks of calculus with the method above, when L'Hopital is not available. – zhw. Oct 16 '15 at 21:07