Evaluate the limit without using L'hopital's rule
a)$$\lim_{x \to 0} \frac {(1+2x)^{1/3}-1}{x} $$
I got the answer as $l=\frac 23$... but I used L'hopitals rule for that... How can I do it another way?
b)$$\lim_{x \to 5^-} \frac {e^x}{(x-5)^3}$$
$l=-\infty$
c)$$\lim_{x \to \frac {\pi} 2} \frac{\sin x}{\cos^2x} - \tan^2 x$$
I don't know how to work with this at all
So basically I was able to find most of the limits through L'Hopitals Rule... BUT how do I find the limits without using his rule?
On
a) Using $$(a-b)(a^2+ab+b^2)=a^3-b^3$$ $a=(1+2x)^{1/3},b=1$ $$\frac{(1+2x)^{1/3}-1}{x}=\frac{(1+2x)^{1/3} - 1}{x}\frac{(1+2x)^{2/3}+(1+2x)^{1/3}+1}{(1+2x)^{2/3}+(1+2x)^{1/3}+1}=$$ $$=\frac{(1+2x) - 1}{x((1+2x)^{2/3}+(1+2x)^{1/3}+1)}=\frac{2x}{x((1+2x)^{2/3}+(1+2x)^{1/3}+1)}=$$ $$=\frac{2}{(1+2x)^{2/3}+(1+2x)^{1/3}+1}$$
On
For small $z$, $(1+z)^n =1+nz+O(z^2) \approx 1+nz $ so $(1+nz)^{1/n} \approx 1+z $ or $(1+z)^{1/n} \approx 1+z/n $.
Therefore $\frac {(1+2x)^{1/3}-1}{x} \approx \frac {(1+2x/3)-1}{x} = \frac {2x/3}{x} =2/3 $.
On
$(a)$ Set $(1+mx)^{\dfrac1n}=1+u\implies1+2m=(1+u)^n$
$$F=\lim_{x\to0}\frac{(1+mx)^{\frac1n}-1}x=\lim_{u\to0}\frac{m(u)}{(1+u)^n-1}$$
$$=m\lim_{u\to0}\frac u{(1+nu+O(u^2))-1}$$
As $u\to0,u\ne0$
$$F=m\frac1{\lim_{u\to0}[n+O(u)]}=\frac mn$$
Here $m=2,n=3$
On
$$L_1=\lim_{x \to 0} \frac {(1+2x)^{1/3}-1}{x}$$
Using binomail thorem: $$(1+2x)^{1/3}=1+\frac13.2x+\text{terms containing other higher powers of x}$$ Putting back in you'll get the limit as $2/3$
$$L_2=\lim_{x \to 5^-} \frac {e^x}{(x-5)^3}=-\infty$$
Because denominator $\to0^-$ and the numerator is finite.
$$L_3=\lim_{x \to \pi/2} \frac{\sin(x)}{\cos(x)^2} - \tan(x)^2$$
Since, $$\frac{\sin(x)}{\cos(x)^2} - \tan(x)^2=\frac{\sin x-\sin^2 x}{\cos^2x}=\frac{\sin x(1-\sin x)}{(1-\sin x)(1+\sin x)}=\frac{\sin x}{1+\sin x}$$ So, obviously limit is $\frac 12$
On
a) $$\lim_{x\to0}{\frac{\big(1+2x\big)^{\frac{1}{3}}-1}{x}}$$ First, let $f(x)=\big(1+2x\big)^{\frac{1}{3}}$, we know that $f(0)=1$ and $f'(x)=\frac{2}{3\big(1+2x\big)^{\frac{2}{3}}}$ $\Rightarrow$ $f'(0)=\frac{2}{3}$
so $$\lim_{x\to0}{\frac{\big(1+2x\big)^{\frac{1}{3}}-1}{x}}$$ $$=\lim_{x\to0}{\frac{f(x)-f(0)}{x-0}}=f'(0)=\frac{2}{3}$$
b) $$\lim_{x\to5^-}{\frac{e^x}{\big(x-5\big)^3}}$$ $$=\lim_{x\to5^-}{e^x}\cdot\lim_{x\to5^-}{\frac{1}{\big(x-5\big)^3}}$$ $$=e^5\cdot-\infty=-\infty$$
c) $$\lim_{x\to\frac{\pi}{2}}{\frac{\sin x}{\cos^2 x}-\tan^2 x}=L$$
Let's multiply by the conjugate of $\frac{\sin x}{\cos^2 x}-\tan^2 x$:
$$L=\lim_{x\to\frac{\pi}{2}}{\frac{\big(\frac{\sin x}{\cos^2 x}-\tan^2 x\big)\big(\frac{\sin x}{\cos^2 x}+\tan^2 x\big)}{\frac{\sin x}{\cos^2 x}+\tan^2 x}}$$
By smplifying the numerator we get:
$\big(\frac{\sin x}{\cos^2 x}-\tan^2 x\big)\big(\frac{\sin x}{\cos^2 x}+\tan^2 x\big)=\frac{\sin^2 x}{\cos^4 x}-\tan^4 x$
$=\tan^2 x\big(\frac{1-\sin^2 x}{\cos^2}\big)=\tan^2 x$
$$\Rightarrow L=\lim_{x\to\frac{\pi}{2}}{\frac{\tan^2 x}{\frac{\sin x}{\cos^2 x}+\tan^2 x}}$$ $$=\lim_{x\to\frac{\pi}{2}}{\frac{\sin x}{1+\sin x}}=\frac{1}{2}$$
On
There are no approximations or the biniomial theorem needed. For a) we use that the derivate of the mapping $f(x):=(1+2x)^\frac{1}{3}$ exists in $\mathbf R_{\geq 0}$. So the limit $$\lim_{x\to 0}\frac{(1+2x)^{\frac{1}{3}}-1}{x}$$ is the difference quotient at $0$. For this reason we get
$$\lim_{x\to 0}\frac{(1+2x)^{\frac{1}{3}}-1}{x}=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=f'(0)=\left.\frac{2}{3(2x+1)^{\frac{2}{3}}}\right|_{x=0}=\frac{2}{3}.$$
For the first one, note how similar it looks to the derivative of a certain function..
For the second, note that $e^x$ is and nonzero at $5$ and doesn't approach $0$ but something rather bad happens to $\dfrac{1}{(x-5)^3}$ at $x=5$. Plot it if you can't tell what is happening.
For the third, try making a common denominator:
$$\frac{\sin x}{\cos^2 x}-\tan^2 x = \frac{\sin x}{\cos^2 x} - \frac{\sin^2 x}{\cos^2 x} = \frac{\sin x -\sin^2 x}{\cos^2 x}.$$
As written, this is still not entirely easy but we can rewrite it as
$$\frac{\sin x-\sin^2 x}{1-\sin^2 x}.$$
Can you see how to proceed?