find the ratio of curved surface area of a two cone if their diameter of the basis are equal and plant height are in the ratio 5ratio7​

Answer:

The entrance would look like a mansion.

[tex]\sf \displaystyle \lim_{x \to 0} \bigg(\frac{e^{x} + sin(x) - 1}{log(1 + x)} \bigg)[/tex]

As we know that,

Put the value of x = 0 and check their indeterminant form, we get.

[tex]\sf \displaystyle \lim_{x \to 0} \bigg(\frac{e^{0} + sin(0) - 1}{log(1 + 0)} \bigg) = \frac{0}{0} \ form[/tex]

As we can see that,

It is 0/0 indeterminant form.

As we know that,

Standard limits.

[tex]\sf \displaystyle e^{x} = 1 + x + \frac{x^{2} }{2!} + \frac{x^{3} }{3!} + . . . . .[/tex]

[tex]\sf \displaystyle sin(x) = x - \frac{x^{3} }{3!} + \frac{x^{5} }{5!} - . . . . .[/tex]

[tex]\sf \displaystyle log(1 + x) = x - \frac{x^{2} }{2} + \frac{x^{3} }{3} - . . . . .[/tex]

Using this standard limits in this question, we get.

[tex]\sf \displaystyle \lim_{x \to 0} \bigg[ \frac{(1 + x + \frac{x^{2} }{2!} + . . . . . )+ (x - \frac{x^{3} }{3!} + \frac{x^{5} }{5!} - . . . . . ) - 1}{(x - \frac{x^{2} }{2}+ \frac{x^{3} }{3} - . . . . . ) } \bigg][/tex]

[tex]\sf \displaystyle \lim_{x \to 0} \bigg[ \frac{1 + x + x - 1}{x } \bigg] = \frac{2x}{x} = 2[/tex]

[tex]\sf \displaystyle \boxed{\lim_{x \to 0} \bigg(\frac{e^{x} + sin(x) - 1}{log(1 + x)} \bigg) = 2}[/tex]

[tex]\sf \displaystyle \lim_{x \to 0} \frac{sin(x)}{x} = \lim_{x \to 0} \frac{x}{sin(x)} = 1 \ \ and \ \ \lim_{x \to 0} sin(x) = 0[/tex]

[tex]\sf \displaystyle \lim_{x \to 0} cos (x) = \lim_{x \to 0} \frac{1}{cos(x)} = 1[/tex]

[tex]\sf \displaystyle \lim_{x \to 0} \frac{tan(x)}{x} = \lim_{x \to 0} \frac{x}{tan(x)} = 1 \ \ and \ \ \lim_{x \to 0} tan(x) = 0[/tex]