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$$\lim_{x \rightarrow \infty}\frac{5^{x+1}+7^{x+1}}{5^x +7^x}$$

I tried using L'Hospital rule, which yielded :

$$\lim_{x \rightarrow \infty}\frac{5^{x+1} \ln 5+7^{x+1} \ln 7}{5^x \ln 5 }$$

But I'm at the dead end... If I divide numerator and denominator by $5^x$ , I get a term $\frac{7^x}{5^x}$ ... which is unsolvable for the limit $x \rightarrow \infty $

However , the answer provided by book is $-7$ . I doubt there is mistake in the question.

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  • $\begingroup$ As written, there is a mistake somewhere. You have strictly positive values everywhere except for a limit that is magically negative. $\endgroup$ – Hans Musgrave Aug 16 '18 at 6:58
  • $\begingroup$ @HansMusgrave.... Is there a way out? Keeping aside the answer of the book. I tried L'Hospital. But failed. I tried to use formula. Failed. $\endgroup$ – Entrepreneur Aug 16 '18 at 7:00
  • $\begingroup$ I like @giannispapav's answer for this kind of thing. Divide by the biggest things in sight and see what's left and what approaches $0$. Otherwise, I personally might have started the problem by breaking the fraction apart to make L'Hopital's result more obvious. $\endgroup$ – Hans Musgrave Aug 16 '18 at 7:02
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The standard method is to factor out and cancel the highest power, which is here $7^x$:

We get:

$\lim_{x\to\infty} \frac{5^{x+1}+7^{x+1}}{5^x+7^x}=\lim_{x\to\infty} \frac{7^{x+1}\left(\left(\frac57\right)^{x+1}+1\right)}{7^x\left(\left(\frac57\right)^x+1\right)}=\lim_{x\to\infty} \frac{7\left(\left(\frac57\right)^{x+1}+1\right)}{\left(\frac57\right)^x+1}$

$=7\cdot\lim_{x\to\infty} \frac{\left(\frac57\right)^{x+1}+1}{\left(\frac57\right)^x+1}$

It is $\lim_{x\to\infty} \left(\frac{5}{7}\right)^x=0$, since $\frac57<1$

We get:

$=7\cdot\frac{0+1}{0+1}=7$

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$\dfrac{5^{x+1}+7^{x+1}}{5^{x}+7^{x}}=\dfrac{5(\frac{5}{7})^x+7}{(\frac{5}{7})^x+1}\to \dfrac{5\cdot0+7}{0+1}=7$

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  • $\begingroup$ Is $\frac{5^x}{7^x}=0$ in the limit $x\rightarrow \infty $ ? $\endgroup$ – Entrepreneur Aug 16 '18 at 7:02
  • $\begingroup$ Not exactly ... $\frac{5^x}{7^x}=(\frac{5}{7})^x\to 0$ as $x\to \infty$ since $\frac{5}{7}<1$ $\endgroup$ – 1123581321 Aug 16 '18 at 7:04
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$$\lim_{x \rightarrow \infty}\frac{5^{x+1}+7^{x+1}}{5^x +7^x}=\lim_{x \rightarrow \infty}\frac{7^{x+1}({5^{x+1}\over 7^{x+1}}+1)}{7^{x}({5^x \over 7^{x}}+1)}$$ $$= 7\lim_{x \rightarrow \infty}\frac{{\big({5\over 7}\big)^{x+1}}+1}{\big({5\over 7}\big)^{x}+1}$$ $$= 7\lim_{t \rightarrow 0}\frac{{5\over 7}t+1}{t+1} =7$$ where $t= \big({5\over 7}\big)^{x}$

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Simplify the fraction by $7^x$, and then everything will have a limit. The limit is $7$, not $-7$, so maybe there is a typo.

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$$\lim_{x\to \infty}\frac{7^{x+1}((\frac57)^{x+1}+1)}{7^x((\frac57)^x+1)}$$

when $x\to \infty$ since $\frac57 \lt1$ then $(\frac57)^{x+1}\to 0$ similar with $(\frac57)^{x}$

Divide $7^{x+1}$ with $7^{x}$ which is 7, therefore, the limit is $7$

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