L'Hospital's Rule application with raised exponents.

Question

I am having a little trouble going about: $$\lim_{x\to \infty} \left(\frac{14x}{14x+10}\right)^{10x}$$

Using $\ln$ properties we can bring down the $10x$ exponent and have:$$\ln y=10x\ln\left(\frac{14x}{14x+10}\right)$$

And from here I get stuck trying to apply L'Hospital's Rule to find the limit.

Answer 1

Hint :

$$10x \ln \left(\frac{14x}{14x+10}\right)= \frac{\ln \left(\frac{14x}{14x+10}\right)}{\frac{1}{10x}}$$

Answer 2

HINT

We can use that

$$\left(\frac{14x}{14x+10}\right)^{10x}=\left(\frac{14x+10-10}{14x+10}\right)^{10x}=\left[\left(1-\frac{10}{14x+10}\right)^{\frac{14x+10}{10}}\right]^{\frac{10\cdot 10x}{14x+10}}$$

and refer to standard limits.

Answer 3

Tips:

With equivalents, it's very fast:$$\ln\biggl(\frac{14x}{14x+10}\biggr)=\ln\biggl(1-\frac{10}{14x+10}\biggr)\sim_\infty -\frac{10}{14x+10}\sim_\infty -\frac{10}{14x}=-\frac 57,$$ so, as equivalence is compatible with multiplication/division, $$10x\ln\biggl(\frac{14x}{14x+10}\biggr)\sim_\infty -\frac{50x}{7x}\to -\frac{50}7.$$ Warning: equivalence is not compatible with addition/subtraction. It is compatible, under mild hypotheses, with composition on the left by logarithm.

Answer 4

Hint: You want to reach some kind of indeterminate form, so you need to mess around with the fractions to reach a desirable form.

$$\ln y=10x\ln\left(\frac{14x}{14x+10}\right)$$

$$\ln y=\frac{\ln\left(\frac{14x}{14x+10}\right)}{\frac{1}{10x}}$$

Now, you can continue.

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As an alternative approach (both faster and easier), you can use $\lim_\limits{n \to \infty}\big(1+\frac{x}{n}\big)^n = e^x$. These expressions are easily manipulated to reach such a form.

Notice the numerator is $10$ less than the denominator, so $\frac{14x}{14x+14} = 1+\frac{-10}{14x+10}$. Hence, you get

$$\lim_{x\to \infty} \left(\frac{14x}{14x+10}\right)^{10x} = \lim_{x\to \infty} \Biggl[\left(1+\frac{-10}{14x+10}\right)^{14x+10}\Biggl]^{\frac{10x}{14x+10}}$$

Answer 5

Once you get to the logarithm, you need $$ \lim_{x\to\infty}\frac{\ln\dfrac{14x}{14x+10}}{\dfrac{1}{10x}} $$ which is in the form $0/0$ and l'Hôpital can be applied. However it's much simpler if the numerator is written as $\ln14+\ln x-\ln(14x+10)$, in order to compute the derivative (you know the limit is $0$ anyhow). So we get $$ \lim_{x\to\infty}\frac{\dfrac{1}{x}-\dfrac{14}{14x+10}}{-\dfrac{1}{10x^2}}= \lim_{x\to\infty}-10x^2\frac{14x+10-14x}{x(14x+10)}=\lim_{x\to\infty}-\frac{100x}{14x+10} $$ When you have this limit, let me call it $l$, the one you started with is $e^l$.